Permutations and Combinations: Class 11 Maths Chapter 6 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

90 min
1 simulation
50 practice questions
21 MCQs

Is chapter mein aap kya seekhenge

Fundamental principle of counting aur alag-alag cases ke liye addition principle lagakar outcomes ko bina list kiye count karna
Factorial notation, jismein 0! = 1 bhi shaamil hai, use karke expressions simplify karna aur equations solve karna
nPr = n!/(n − r)! aur n^r derive karna aur unse distinct objects ke arrangements count karna, bina repetition aur repetition ke saath
Jab kuch objects ek jaise hon tab arrangements count karna, aur fixed places, saath rakhe blocks aur alag rakhe objects wale cases sambhalna
nPr = nCr × r! se nCr = n!/(r!(n − r)!) derive karna, aur symmetry property aur Pascal's rule use karna
Pehchanna ki problem mein arrangements chahiye ya selections, aur board aur entrance level ke mixed problems solve karna

NCERT Mathematics Class 11, Chapter 6, pp. 100–125 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

6.1 Introduction

India mein kisi bhi car ya scooter ki number plate dekho. State code aur district number ke baad letters ki ek chhoti series aati hai aur phir chaar digits, jaise MH 12 KX 4071 mein. Maan lo ek district is tail ko koi bhi do letters aur uske baad koi bhi chaar digits se bhar sakta hai. Khatam hone se pehle woh kitni alag tails issue kar sakta hai?

Inhe list karne mein hafte lag jaayenge; answer hai $26 \times 26 \times 10^{4}$ , yaani 67.6 lakh. Yeh chapter aise numbers tak list karke nahi, reasoning se pahunchta hai. Do sawaal poore chapter mein chalte hain: objects ko order mein kitne tareekon se arrange kiya ja sakta hai, aur jab order matter nahi karta to ek group kitne tareekon se select kiya ja sakta hai? Pehla sawaal permutations tak le jaata hai, doosra combinations tak. Dono ek simple rule par tike hain, jo agle section ka topic hai.

6.2 Fundamental Principle of Counting

School canteen do tarah ke paratha banati hai, aloo aur gobhi, aur teen drinks deti hai: lassi, chai aur chaas. Ek meal matlab ek paratha aur ek drink. Tum koi bhi paratha lo, teeno drinks phir bhi milti hain, isliye har paratha se 3 meals bante hain aur do parathon se $2 \times 3 = 6$ meals. Tree diagram isse saaf dikhata hai (Fig. 6.1): top se kisi leaf tak har path ek meal hai.

3/3Chhe leaves, har meal ke liye ek: $2 \times 3 = 6$ .

Fig. 6.1Pehle 2 choices mein se har ek ke baad utne hi second choices khulte hain, 3, isliye 6 paths bante hain.NCERT Fig. 6.1 · p. 101

Upar se neeche bana ek tree. Pehle step mein ek start point do boxes mein bantta hai, aloo aur gobhi, paratha ki choice ke liye. Doosre step mein har paratha box teen dashed boxes mein bantta hai, lassi, chai aur chaas, to neeche chhe leaves dikhti hain. Teesre step mein chhe leaves par 1 se 6 tak number lagte hain aur neeche line aati hai: 2 times 3 equals 6 meals.

Ab ek stage aur jodo. Shaadi ki thali mein 3 mein se ek sabzi, 2 mein se ek dal aur 4 mein se ek sweet hai. $3 \times 2 = 6$ sabzi–dal pairs mein se har ek ko 4 sweets mein se kisi se bhi poora kiya ja sakta hai, isliye $6 \times 4 = 24$ alag thalis bante hain. Woh tree banane mein 24 leaves chahiye; multiply karne mein sirf ek line. Yahi rule fundamental principle of counting hai; kai books ise multiplication principle kehti hain.

PrincipleFundamental principle of counting

Do-stage process ko stage by stage count karo. Agar stage one ke $m$ possible outcomes hain, aur unmein se har ek ke liye stage two ke $n$ possible outcomes hain, to poore process ke $m \times n$ possible outcomes hote hain. Fixed order mein aur stages hon to multiply karte jao.

Multiplication principle

N = m_{1} m_{2} \dots m_{k}

(6.1)

jahan

$m_{i}$: task $i$ karne ke kitne tareeke hain, jab tasks 1 se $i - 1$ ho chuke hon
$N$: saare $k$ tasks ko order mein karne ke kitne tareeke hain

Example 6.1NCERT Example 1 · p. 102

Ek row mein paanch letters

PLANT ke letters se kitne five-letter strings (zaroori nahi ki words hon) ban sakte hain agar koi letter do baar use na ho? Aur agar letters repeat ho sakte hain to kitne?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.2NCERT Example 2 · p. 103

Rangoli ke do rings

Ek rangoli mein ek inner ring aur ek outer ring hai. Har ring 5 colours mein se chune gaye ek colour se bhari jaati hai, aur dono rings ke colours alag hone chahiye. Kitni colourings possible hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.3NCERT Example 3 · p. 103

Odd numbers, digits repeat ho sakte hain

Digits 2, 3, 4, 5, 6, 7 se kitne three-digit odd numbers ban sakte hain agar digits repeat ho sakte hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.4NCERT Example 4 · p. 103

Alag lengths ke codes

Ek club ka door code A, B, C, D, E, F mein se liye gaye 2, 3 ya 4 alag letters ki row hai. Kitne codes possible hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Counting probability ka bhi pehla step hai, jo tum Chapter 14 mein padhoge. Do dice ek saath phenkne par $6 \times 6 = 36$ ordered outcomes milte hain, kyunki pehle die ke 6 faces mein se har ek doosre die ke 6 faces ke saath pair banata hai. Inmein se sirf 6 outcomes doubles hain, isliye double lagbhag chhe throws mein ek baar aata hai. Neeche wala lab tumhe yeh count test karne deta hai.

SimulationDo dice: 6 × 6 outcomes

Do dice ka ek lab. Buttons pair ko ek baar, 10, 100 ya 1000 baar roll karte hain. Ek bar chart dikhata hai ki 2 se 12 tak ka har total kitni baar aaya, uske theoretical share ke saath, aur ek readout doubles ke fraction ko one sixth ki theoretical value se compare karta hai. Jab tak learner roll button nahi dabata, kuch nahi hilta.

Fig. 6.2Har throw

6 \times 6 = 36

ordered pairs mein se ek hai, aur unmein se 6 doubles hain. Counts ko theory se compare karne ke liye roll button dabao.

Try karo

Roll karne se pehle: do dice ke kitne ordered outcomes hote hain, aur unmein se kitne doubles hain?
1000 baar roll karo. Kya doubles ka share $\frac{6}{36} = \frac{1}{6}$ ke kareeb hai?
Total 7, total 2 se zyada baar kyun aata hai? Har ek dene wale ordered pairs gino.

Practice · 7 questions

Practice 6.1: answers ke saath

1
Short answerEasyNCERT Exercise 6.1, Q1
Sirf digits 2, 4, 5, 7, 8 aur 9 allowed hain. Kitne 4-digit numbers likh sakte ho agar
1. digits repeat ho sakte hain?
  Answer dekho
  Answer 1296
  Har place ke 6 choices hain: $6^{4} = 1296$ .
2. koi digit repeat nahi ho sakta?
  Answer dekho
  Answer 360
  $6 \times 5 \times 4 \times 3 = 360$ .
Short answerMediumNCERT Exercise 6.1, Q2
Digits repeat ho sakte hain. Sirf 1, 2, 3, 4, 5, 6 aur 7 use karke kitne 4-digit multiples of 5 likh sakte ho?
Answer dekho
Answer 343
0 available nahi hai, isliye units digit 5 hi hona chahiye: 1 tareeka. Baaki teen places 7 mein se koi bhi digit le sakti hain: $7^{3} = 343$ .
Short answerEasyNCERT Exercise 6.1, Q3
O se Z tak ke 12 letters se kitne 3-letter codes ban sakte hain, jismein koi letter do baar use na ho?
Answer dekho
Answer 1320
$12 \times 11 \times 10 = 1320.$
Short answerMediumNCERT Exercise 6.1, Q4
Ek 6-digit locker code 40 se shuru hona chahiye, aur koi digit do baar nahi aa sakta. Kitne codes possible hain?
Answer dekho
Answer 1680
Chaar places ke liye aath digits bachte hain: $8 \times 7 \times 6 \times 5 = 1680.$
Short answerEasyNCERT Exercise 6.1, Q5
Ek die ek baar roll kiya jaata hai aur phir ek coin do baar toss kiya jaata hai. Teeno results ke kitne alag records possible hain?
Answer dekho
Answer 24
$6 \times 2 \times 2 = 24$ .
Short answerEasyNCERT Exercise 6.1, Q6
Alag-alag colours ki saat lanterns available hain. Ek festival signal unmein se 3 ko ek ke neeche ek latkata hai. Kitne signals possible hain?
Answer dekho
Answer 210
$7 \times 6 \times 5 = 210$ . Order matter karta hai kyunki upar se neeche ka order signal ka hissa hai.
Quick answerHard
Kitne 3-digit numbers mein kam se kam ek digit repeat hota hai?
252648900271
Answer dekho
Answer (A) 252
Kul 900 three-digit numbers hain. Jinke saare digits alag hain, unki ginti $9 \times 9 \times 8 = 648$ hai: pehla digit 0 nahi ho sakta, doosra ho sakta hai. To $900 - 648 = 252$ mein repeat hai.

6.3 Permutations

Shelf par do books ki jagah badlo aur shelf alag dikhti hai, halaanki wahi books wahan hain. Jab bhi objects ka order us cheez ka hissa ho jo hum gin rahe hain, har possible order ko permutation (क्रमचय) kehte hain. Example 6.1 ke strings PLANT ke letters ke permutations the.

Definition

Permutation

Objects ki ek ordering: unmein se kuch ya saare ek ke baad ek rakhe jaate hain, aur order badalne se alag permutation banta hai.

ExamplePLANET ke letters se LANE aur ELAN 4 letters ke do alag permutations hain.

Permutation sirf kuch objects bhi use kar sakta hai. PLANET ke letters se bana 4-letter code, jismein koi letter repeat na ho, 6 letters ka 4 at a time permutation hai, aur slot method se aise $6 \times 5 \times 4 \times 3 = 360$ codes milte hain. Agle subsections aise products ko formulas mein badalte hain.

6.3.1 Jab saare objects distinct hon tab permutations

$n$ alag objects aur ek row mein $r$ khaali places lo, jahan $0 < r \leq n$ . Places ko ek-ek karke bharo aur kisi object ko dobara use mat karo. Fig. 6.3 yahi chhe letter tiles aur teen places ke saath karta hai.

4/44 tiles bache hain, isliye $6 \times 5 \times 4 = 120$ arrangements hain.

Fig. 6.3Har use hua tile agli place ke liye ek kam chhodta hai.

Upar teen dashed boxes hain aur neeche ek row mein chhe letter tiles, A se F, rakhe hain. Doosre step mein tile C upar pehle box mein jaata hai aur uske neeche label 6 ways aata hai. Teesre step mein tile F doosre box mein jaata hai, label 5 ways, kyunki sirf paanch tiles bache the. Chauthe step mein tile A teesra box bharta hai, label 4 ways, aur neeche product 6 times 5 times 4 equals 120 dikhta hai. Dashed outlines dikhati hain ki use hue tiles kahan the.

Theorem 6.1Distinct objects ke arrangements

Maan lo $0 < r \leq n$ . Agar koi object do baar use nahi ho sakta, to $n$ alag objects mein se liye gaye $r$ objects ko $n (n - 1) (n - 2) \dots (n - r + 1)$ tareekon se line mein lagaya ja sakta hai. Is number ko $^{n} P_{r}$ likhte hain.

Product mein $r$ factors hain, jo $r$ bada hone par bhaari pad jaata hai. Ek naya notation isse theek karta hai.

6.3.2 Factorial notation

Natural number $n$ ke liye, 1 se $n$ tak ke saare natural numbers ke product ko $n!$ likhte hain aur $n$ factorial (क्रमगुणित) padhte hain. To $1! = 1$ , $2! = 2$ , $3! = 6$ , $4! = 24$ aur $5! = 120$ . Factorials bahut tezi se badhte hain: $10!$ pehle hi 36 lakh se zyada hai.

Definition

Factorial

Natural number $n$ ke liye, $n! = 1 \times 2 \times 3 \times \dots \times n .$ Agreement se, $0! = 1$ .

Example $6! = 720$

Sabse bada factor alag karne par ek chhota factorial bachta hai: $6! = 6 \times 5!$ aur $5! = 5 \times 4!$ . Yeh step dohrane par $n \geq 2$ ke liye $n! = n (n - 1) (n - 2)!$ milta hai, aur aise hi aage. Basic step yeh hai:

Factorial, ek step neeche

n! = n \times (n - 1)!

(6.2)

$0!$ ki value 1 kyun honi chahiye? (6.2) ko ulta padho: har step neeche jaane par us number se divide hota hai jo peeche chhoot jaata hai. Fig. 6.4 $4!$ se neeche utarta hai. Aakhri step, $1!$ se $0!$ tak, 1 ko 1 se divide karta hai aur 1 bachta hai. $0! = 1$ maan lene se pattern nahi tootta, aur isse kai formulas apne edge cases par bhi kaam karte hain.

3/31 se divide karo: pattern kehta hai $0! = 1$ .

Fig. 6.4Ek step neeche jaane par us number se divide hota hai jo peeche chhoot jaata hai.

Boxes ki ek seedhi upar left se neeche right ki taraf utarti hai. Pehle step mein teen boxes 4! equals 24, 3! equals 6 aur 2! equals 2 dikhate hain, jo divide by 4 aur divide by 3 label wale curved arrows se jude hain. Doosre step mein chautha box aata hai, 1! equals 1, jo 2 se divide karke milta hai. Teesre step mein ek highlighted aakhri box aata hai, 0! equals 1, jo 1 ko 1 se divide karke milta hai.

Example 6.5NCERT Example 5 · p. 106

Factorials ki value nikalna

Value nikalo (i) $6!$ (ii) $8!$ (iii) $8! - 6!$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.6NCERT Example 6 · p. 106

Factorials cancel karna

Compute karo (i) $\frac{9!}{6!}$ (ii) $\frac{15!}{13! 2!}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.7NCERT Example 7 · p. 106

Factorial expression mein values daalna

Jab $n = 7$ aur $r = 3$ ho to $\frac{n!}{r! (n - r)!}$ ki value nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.8NCERT Example 8 · p. 106

Factorial equation mein ek unknown

$x$ nikalo agar $\frac{1}{7!} + \frac{1}{8!} = \frac{x}{9!}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

\frac{10!}{8!}

kiske barabar hai?

\frac{10}{8}

90

2!

80

Answer dekho

Answer (B) $90$

10! = 10 \times 9 \times 8!

, isliye quotient

10 \times 9 = 90

hai. 10 ko 8 se divide karna factorial sign ka matlab ignore kar deta hai.

Practice · 5 questions

Practice 6.2: answers ke saath

1
Short answerEasyNCERT Exercise 6.2, Q1
Value nikalo
1. $7!$
  Answer dekho
  Answer 5040
  $7! = 720 \times 7 = 5040$ .
2. $5! - 4!$
  Answer dekho
  Answer 96
  $4! (5 - 1) = 24 \times 4 = 96.$
Quick answerEasyNCERT Exercise 6.2, Q2
True ya false: $(2 \times 3)! = 2! \times 3!$ .
SahiGalat
Answer dekho
Answer Galat
$(2 \times 3)! = 6! = 720$ , jabki $2! \times 3! = 2 \times 6 = 12$ .
Short answerEasyNCERT Exercise 6.2, Q3
$\frac{10!}{7! 3!}$ compute karo.
Answer dekho
Answer 120
$\frac{10 \times 9 \times 8}{6} = 120$ .
Short answerMediumNCERT Exercise 6.2, Q4
$x$ nikalo agar $\frac{1}{4!} - \frac{1}{5!} = \frac{x}{6!}$ .
Answer dekho
Answer 24
$\frac{6!}{4!} - \frac{6!}{5!} = 30 - 6 = 24.$
5
Short answerEasyNCERT Exercise 6.2, Q5
$\frac{n!}{(n - r)!}$ ki value nikalo jab
1. $n = 8$ , $r = 3$
  Answer dekho
  Answer 336
  $8 \times 7 \times 6 = 336$ .
2. $n = 10$ , $r = 4$
  Answer dekho
  Answer 5040
  $10 \times 9 \times 8 \times 7 = 5040.$

6.3.3 nPr ke formula ka derivation

DerivationLambe product se factorials tak

Theorem 6.1 se shuru karo: $^{n} P_{r}$ , $n$ se $n - r + 1$ tak ke $r$ numbers ka product hai. Unke neeche ke numbers, $n - r$ se 1 tak, missing hain. Saath milkar woh $(n - r)!$ banate hain.

Is missing tail se multiply aur divide karo. Numerator poora product $n!$ ban jaata hai, isliye $0 \leq r \leq n$ ke liye:

Distinct objects ke permutations

^{n} P_{r} = \frac{n!}{(n - r)!}

(6.3)

Fig. 6.5 yeh step $n = 7$ aur $r = 3$ ke saath dikhata hai.

3/3Extra factors cancel ho jaate hain, isliye $^{7} P_{3} = \frac{7!}{4!} = 210$ .

Fig. 6.5Missing tail se multiply aur divide karne par product factorials ka quotient ban jaata hai.

Pehle step mein teen solid boxes mein 7, 6 aur 5 hain, jo 7P3 ke teen factors hain. Doosre step mein chaar dashed boxes, 4, 3, 2 aur 1, row mein judkar 7! banate hain, ek fraction bar aata hai, aur wahi chaar dashed boxes bar ke neeche 4! ke roop mein aate hain. Teesre step mein tirchhi lines bar ke upar aur neeche ke dashed boxes ko kaat deti hain, jo fade ho jaate hain, aur result 7 times 6 times 5 equals 210 dikhta hai.

Do edge cases muft mein mil jaate hain. $r = n$ par, $^{n} P_{n} = \frac{n!}{0!} = n!$ : saare $n$ objects ek row mein. $r = 0$ par formula $\frac{n!}{n!} = 1$ deta hai, jo theek baithta hai: khaali row hi aisi akeli row hai jismein koi object nahi. To (6.3) 0 se $n$ tak har $r$ ke liye sahi hai.

Theorem 6.2Repetition ke saath arrangements

Jab objects ko jitni baar chahe dobara use kar sakte hain, to $n$ alag objects se $r$ places $n^{r}$ tareekon se bhari ja sakti hain.

Formulas hamare slot counts ko ek line mein dobara kar dete hain: Example 6.1 ke PLANT strings ke liye $^{5} P_{5} = 5! = 120$ aur $5^{5} = 3125$ , aur PLANET codes ke liye $^{6} P_{4} = \frac{6!}{2!} = 360$ . Naye cases bhi utni hi jaldi solve hote hain. Agar 10 passengers 3 alag berths, lower, middle aur upper, ka wait kar rahe hain, to berths $^{10} P_{3} = 720$ tareekon se di ja sakti hain; order matter karta hai kyunki berths alag hain.

Quick check

2 mein se 1

Kaunsa expression

^{9} P_{4}

ke barabar hai?

\frac{9!}{4!}

9^{4}

\frac{9!}{4! 5!}

\frac{9!}{5!}

Answer dekho

Answer (D) $\frac{9!}{5!}$

^{9} P_{4} = \frac{9!}{(9 - 4)!} = \frac{9!}{5!} = 3024.

4!

se divide karna wahan

r

use karta hai jahan

n - r

aana chahiye.

6.3.4 Jab saare objects distinct na hon tab permutations

Ab tak har object alag tha. Jab kuch ek jaise hon to kya badalta hai? Word MAA lo, jiska Hindi mein matlab mother hai, jismein do identical A hain. Thodi der ke liye unhe $A_{1}$ aur $A_{2}$ tag karo, taaki teeno letters alag ho jaayein; tab $3! = 6$ arrangements hain. Ab tags hata do. $M A_{1} A_{2}$ aur $M A_{2} A_{1}$ dono MAA ban jaate hain, aur har untagged word theek $2! = 2$ tagged words se aata hai (Fig. 6.6). To MAA ke $\frac{6}{2} = 3$ arrangements hain.

3/3Har pair ek word ban jaata hai: $\frac{3!}{2!} = 3$ words.

Fig. 6.6Tags hatane par woh orders mil jaate hain jo sirf identical letters ki adla-badli se alag hain.

Chhe boxes ka ek column tagged orders M A1 A2, M A2 A1, A1 M A2, A2 M A1, A1 A2 M aur A2 A1 M ko heading tagged: 3! equals 6 ke neeche list karta hai. Doosre step mein curly braces un boxes ko pairs mein jodte hain jo sirf A1 aur A2 ki adla-badli se alag hain. Teesre step mein har pair right side ke ek untagged word ki taraf point karta hai, MAA, AMA aur AAM, heading tags removed ke neeche, aur uske neeche 6 divided by 2! equals 3 likha hai.

Yahi reasoning bade cases par bhi chalti hai. BANANA mein 6 letters hain: teen A, do N aur ek B. Tag karne par $6!$ orders milte hain. Ek untagged word, A tags ko shuffle karne ke $3!$ tareekon aur alag se N tags ko shuffle karne ke $2!$ tareekon se banta hai, yaani $3! \times 2!$ tagged orders se. Isliye BANANA ke $\frac{6!}{3! 2!} = 60$ arrangements hain.

Theorem 6.3Ek kind repeat hua

Maan lo $n$ objects mein se $p$ identical hain aur baaki $n - p$ unse aur aapas mein alag hain. Tab saare $n$ ko ek row mein $\frac{n!}{p!}$ distinguishable tareekon se rakha ja sakta hai.

Theorem 6.4Kai kinds repeat hue

$n$ objects ko kinds mein baanto: ek kind ke $p_{1}$ identical objects, doosri ke $p_{2}$ , aur aise hi $k$ th kind ke $p_{k}$ tak, aur bache hue saare objects alag. Tab saare $n$ objects use karne wali rows $\frac{n!}{p_{1}! p_{2}! \dots p_{k}!}$ distinguishable tareekon se ban sakti hain.

InteractiveWord arrangement lab

Ek word lab. Preset buttons MAA, BANANA, PAPER aur TIRUPATI load karte hain, aur ek text box 12 letters tak ka koi bhi word leta hai. Letter tiles word dikhate hain, jismein repeated letters dashed hain aur unpar 1, 2, 3 tag lage hain. Ek status panel letter counts, arrangements ki number (n factorial divided by repeat counts ke factorials) aur dictionary order mein word ka rank list karta hai. Find it button wala ek number box dictionary list mein kisi bhi position par aane wala word dikhata hai. Jab 360 ya usse kam arrangements hon, to saare dictionary order mein list hote hain aur type kiya hua word outlined hota hai. Jab tak learner kuch nahi karta, kuch nahi hilta.

Fig. 6.7Ek word type karo, ya koi preset tap karo, aur dekho uske letter counts, arrangements ki number aur dictionary order mein word kahan aata hai.

Example 6.9NCERT Example 9 · p. 110

Ek temple town ke letters

TIRUPATI ke saare letters ke kitne arrangements hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.10NCERT Example 10 · p. 110

Bina repeat ke five-digit numbers

Digits 1 se 8 mein se har ek ko zyada se zyada ek baar use karke kitne 5-digit numbers bana sakte ho?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.11NCERT Example 11 · p. 110

Jab 0 shuru mein nahi aa sakta

0 se 6 tak ke saat digits mein se har ek ko zyada se zyada ek baar use karke kitne 4-digit numbers bana sakte ho?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.12NCERT Example 12 · p. 111

n ke liye solve karna

$n$ nikalo agar (i) $^{n} P_{4} = 20^{n} P_{2}$ ; (ii) $^{n} P_{3} :^{n - 1} P_{3} = 4 : 3$ . Dono mein $n \geq 4$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.13NCERT Example 13 · p. 112

r ke liye solve karna, aur ek root reject karna

$r$ nikalo agar $7 \cdot^{6} P_{r} = 30 \cdot^{7} P_{r - 1}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Yeh letters saath rehne chahiye jaisi conditions gluing se handle hoti hain: group ko ek block maano, blocks arrange karo, phir block ke andar letters arrange karo. Saare saath nahi ke liye har arrangement gino aur glued wale subtract karo. Fig. 6.8 gluing step dikhata hai.

3/37 units ko $7!$ tareekon se aur block ke andar vowels ko $3!$ tareekon se arrange karo.

Fig. 6.8Group ko ek block mein glue karo, units arrange karo, phir block ke andar arrange karo.

Nau letter tiles ki ek row TRIANGLES likhti hai, jismein vowel tiles I, A aur E dashed bane hain. Doosre step mein saare tiles ek doosri row mein aa jaate hain: teeno vowels left mein ek heavy outline ke andar khisak kar ek block banate hain, aur consonants T, R, N, G, L, S unke baad aate hain, kul 7 units. Teesre step mein row ke neeche ek bracket par likha hai 7 units: 7! orders, block ke upar ek arc par likha hai 3! inside, aur neeche product 7! times 3! equals 30240 dikhta hai.

Example 6.14NCERT Example 14 · p. 112

Vowels saath, aur saare saath nahi

TRIANGLES mein 9 alag letters hain. Uske woh arrangements gino jinmein (i) vowels I, A, E saare saath hon; (ii) vowels saare saath na hon.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.15NCERT Example 15 · p. 112

Phoolon ka toran

Darwaze ke ek toran mein 10 phool ek line mein piroye hain: 5 marigolds, 2 roses aur 3 jasmines. Ek kind ke phool identical dikhte hain. Kitne alag patterns possible hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.16NCERT Example 16 · p. 113

ASSESSMENT par chaar sawaal

ASSESSMENT mein 10 letters hain: S chaar baar, E do baar, aur A, M, N, T ek-ek baar. Iske saare letters ke arrangements ki number nikalo, aur unmein se kitne (i) M se shuru hote hain; (ii) jinmein saare vowels saath hain; (iii) jinmein saare vowels saath nahi hain; (iv) A se shuru aur T par khatam hote hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Word LEVEL ke kitne arrangements hain?

120603020

Answer dekho

Answer (C) 30

L aur E dono do-do baar aate hain:

\frac{5!}{2! 2!} = 30

. Sirf ek

2!

se divide karne par 60 aata hai.

Practice · 13 questions

Practice 6.3: answers ke saath

Short answerEasyNCERT Exercise 6.3, Q1
Digits 1 se 7 mein se har ek ko zyada se zyada ek baar use karke kitne 4-digit numbers possible hain?
Answer dekho
Answer 840
$^{7} P_{4} = 7 \times 6 \times 5 \times 4 = 840.$
Short answerMediumNCERT Exercise 6.3, Q2
Kitne 3-digit numbers mein koi digit repeat nahi hota?
Answer dekho
Answer 648
Hundreds: 9 choices (1 se 9). Tens: 9 (ab 0 allowed hai, lekin hundreds wala digit use ho chuka hai). Units: 8. $9 \times 9 \times 8 = 648$ .
Short answerMediumNCERT Exercise 6.3, Q3
2, 3, 4, 5, 6, 8, 9 se kitne 3-digit odd numbers ban sakte hain agar koi digit repeat na ho?
Answer dekho
Answer 90
Units 3, 5, 9 mein se: 3 tareeke; phir hundreds place ke liye 6 digits aur tens ke liye 5: $3 \times 6 \times 5 = 90$ .
4
Short answerEasyNCERT Exercise 6.3, Q4
Digits 1 se 6 ko bina repetition ke use karke, number nikalo
1. kitne 3-digit numbers ban sakte hain;
  Answer dekho
  Answer 120
  $^{6} P_{3} = 120$ .
2. un 3-digit numbers ki jo 400 se bade hain.
  Answer dekho
  Answer 60
  Hundreds digit 4, 5 ya 6: 3 tareeke, phir baaki ke liye $5 \times 4$ : 60.
Short answerEasyNCERT Exercise 6.3, Q5
11 players ke cricket squad ko ek captain aur ek vice-captain chunna hai, do alag players. Kitne choices hain?
Answer dekho
Answer 110
$^{11} P_{2} = 11 \times 10 = 110$ .
Short answerMediumNCERT Exercise 6.3, Q6
$n$ nikalo agar $^{n} P_{2} :^{n + 1} P_{3} = 1 : 12$ .
Answer dekho
Answer 11
$^{n + 1} P_{3} = (n + 1) n (n - 1)$ , isliye ratio $\frac{1}{n + 1} = \frac{1}{12}$ hai aur $n = 11$ .
7
Short answerMediumNCERT Exercise 6.3, Q7
$r$ nikalo agar
1. $^{4} P_{r} = 2^{4} P_{r - 1}$
  Answer dekho
  Answer 3
  $\frac{4!}{(4 - r)!} = \frac{2 \cdot 4!}{(5 - r)!}$ se $5 - r = 2$ milta hai, isliye $r = 3$ .
2. $5^{4} P_{r} = 12^{5} P_{r - 1}$
  Answer dekho
  Answer 2
  Cancel karne par $(6 - r) (5 - r) = 12$ milta hai, to $r^{2} - 11 r + 18 = 0$ aur $r = 2$ ya $9$ . Kyunki $r \leq 4$ , $r = 2$ .
Short answerEasyNCERT Exercise 6.3, Q8
JUPITER ke saare letters ko, har letter ek baar use karke, kitne arrangements bante hain?
Answer dekho
Answer 5040
Saat alag letters: $7! = 5040$ .
9
Short answerMediumNCERT Exercise 6.3, Q9
BIRTHDAY mein 8 alag letters hain (Y ko consonant maano). Isse bane letters ke woh strings gino, koi letter do baar use na ho, jo use karte hain
1. ek baar mein 3 letters?
  Answer dekho
  Answer 336
  $^{8} P_{3} = 336$ .
2. saare 8 letters?
  Answer dekho
  Answer 40320
  $8! = 40 320$ .
3. saare 8 letters, dono ends par ek vowel ke saath?
  Answer dekho
  Answer 1440
  Vowels I aur A dono ends ko $2!$ tareekon se lete hain; 6 consonants beech ko $6!$ tareekon se bharte hain: $2 \times 720 = 1440$ .
Short answerHardNCERT Exercise 6.3, Q10
BOOKKEEPER ke kitne distinct arrangements mein teeno E saare saath nahi hain?
Answer dekho
Answer 141120
Total: $\frac{10!}{2! 2! 3!} = 151 200$ (O do baar, K do baar, E teen baar). EEE ko glue karo: 8 units jismein O aur K do-do baar, $\frac{8!}{2! 2!} = 10 080$ . To $151 200 - 10 080 = 141 120.$
11
Short answerHardNCERT Exercise 6.3, Q11
DEMOCRATS mein 9 alag letters hain. Iske kitne arrangements mein
1. D pehle aur S aakhir mein hai?
  Answer dekho
  Answer 5040
  Baaki 7 letters beech ko bharte hain: $7! = 5040$ .
2. vowels E, O, A saare saath hain?
  Answer dekho
  Answer 30240
  Block aur 6 consonants = 7 units: $7! \times 3! = 30 240$ .
3. D aur S ke beech theek 3 letters hain?
  Answer dekho
  Answer 50400
  D aur S positions (1, 5), (2, 6), …, (5, 9) par aa sakte hain: 5 pairs, 2 orders mein. Baaki 7 letters bachi jagah ko $7!$ tareekon se bharte hain: $5 \times 2 \times 5040 = 50 400.$
Quick answerMedium
Assertion (A): NOON ke letters 6 tareekon se arrange ho sakte hain. Reason (R): $n$ objects, jinmein se $p$ ek kind ke alike hain aur $q$ doosri kind ke alike hain, $\frac{n!}{p! q!}$ tareekon se arrange ho sakte hain.
A aur R dono true hain, aur R, A ko explain karta haiA aur R dono true hain, lekin R, A ko explain nahi kartaA true hai, lekin R false haiA false hai, lekin R true hai
Answer dekho
Answer (A) A aur R dono true hain, aur R, A ko explain karta hai
$\frac{4!}{2! 2!} = 6$ , jo theek do N aur do O par lagaya gaya R hai. Dono true hain aur R, A ko explain karta hai.
Quick answerHard
PERMIT ke letters ke kitne arrangements mein E, I se kahin pehle aata hai?
720240360120
Answer dekho
Answer (C) 360
E aur I ki adla-badli $6! = 720$ arrangements ke pairs banati hai, jismein har pair mein theek ek mein E pehle hai. Isliye answer $\frac{720}{2} = 360$ hai.

6.4 Combinations

Chaar dost, Ira, Jai, Kabir aur Lata, ko apne mein se do ko inter-school quiz mein bhejna hai. Matter yeh karta hai ki kaunsa pair jaata hai; kiska naam pehle bola gaya, isse kisi ko farak nahi padta. Jab sirf yeh matter karta hai ki kaunse objects chune gaye, chunne ka order nahi, tab hum combinations (संचय) gin rahe hote hain.

Definition

Combination

Ek collection se objects ki aisi choice jismein sirf yeh ginta hai ki kaunse objects chune gaye, chunne ka order nahi.

ExampleIra, Jai, Kabir aur Lata mein se pair Ira aur Jai ek combination hai; Jai aur Ira wahi combination hai.

Agar dono ke kaam alag hote, jaise captain aur deputy, to order matter karta aur $^{4} P_{2} = 12$ choices hote. Simple pairs ke roop mein sirf 6 hain: IJ, IK, IL, JK, JL aur KL. 12 mein har pair do baar gina gaya tha, har order mein ek baar. $n$ alag objects mein se $r$ objects ke saare selections ki ginti ko hum $^{n} C_{r}$ likhte hain, to yahan $^{4} C_{2} = 6$ .

Combinations kai jagah chhupe hote hain. Ek league mein jahan 8 teams mein se har team baaki har team se ek baar khelti hai, ek match teams ka ek pair hai, isliye $^{8} C_{2}$ matches hain. Circle par chhe points, jinmein koi teen ek line mein nahi, har pair of points ke liye ek chord aur teen points ke har set ke liye ek triangle dete hain (Fig. 6.9).

2/3Har chord points ke ek pair ko jodti hai: har 2-point selection ke liye ek chord.

Fig. 6.9Chord ek 2-point selection hai aur triangle ek 3-point selection: dono mein order matter nahi karta.NCERT Fig. 6.3 · p. 115

Ek circle par chhe labelled points hain, A se F, barabar doori par; paas mein likha hai six points, no three in a line. Doosre step mein points ko pairs mein jodne wali saari pandrah chords bani hain, aur side text kehta hai a chord is a pair of points, saath mein 6C2 equals 15. Teesre step mein chords fade ho jaati hain aur triangle ACE shaded hota hai, text ke saath a triangle is a set of 3 points aur 6C3 equals 20.

Formula nikalne ke liye selections ko arrangements se compare karo. Fig. 6.10 paanch players P, Q, R, S, T se har captain–vice-captain pair list karta hai: $^{5} P_{2} = 20$ ordered pairs. Har do logon ki team do baar aati hai, PQ aur QP ke roop mein, isliye $\frac{20}{2!} = 10$ teams hain.

3/3Grid ka ek triangle rakho: $\frac{20}{2!} = 10$ teams.

Fig. 6.10Do logon ki har team ordered pairs mein do baar dikhti hai, har order mein ek baar.

Ek grid jismein rows captain ke liye aur columns vice-captain ke liye hain, dono par label P, Q, R, S, T. Diagonal cells grey hain kyunki ek insaan dono posts nahi le sakta, to PQ aur QP jaise 20 cells bachte hain; side text kehta hai 20 ordered pairs, 5P2 equals 20. Doosre step mein cells PQ aur QP highlight hote hain aur ek dashed line diagonal mark karti hai: dono same team hain. Teesre step mein diagonal ke neeche ke cells fade ho jaate hain, aur 10 teams bachti hain, 5C2 equals 10.

Teen logon ke groups bhi aise hi behave karte hain: har 3-person group $3!$ orders mein line mein lag sakta hai, isliye arrangements groups se $3! = 6$ guna zyada hain. General roop mein:

Theorem 6.5Arrangements aur selections

$0 < r \leq n$ ke liye, $^{n} P_{r} =^{n} C_{r} \times r!$ .

$r!$ se divide karke aur (6.3) use karke combinations ka formula milta hai:

Combinations

^{n} C_{r} = \frac{n!}{r! (n - r)!}

(6.5)

InteractiveArrange ya choose lab

Teen mode buttons wala ek counting lab: bina repeats ke arrange, repeats allowed ke saath arrange, aur ek group choose. Do sliders objects ki number n set karte hain, 2 se 6 letters A se F tak, aur places ki number ya group size r. Boxes ki ek row har place ke choices ki number dikhati hai, aur ek status panel count ko uske formula ke saath deta hai: nPr ek falling product ke roop mein, n to the power r, ya nCr jo nPr divided by r factorial hai. Neeche har outcome ek chip ke roop mein list hota hai, 120 tak. Choose mode mein ek checkbox har group ko uske andar ke r factorial arrangements mein khol deta hai. Jab tak learner kuch nahi karta, kuch nahi hilta.

Fig. 6.11Ek mode chuno aur

n

aur

r

set karo. Choose a group mein box tick karo taaki har selection apne

r!

orders mein khul jaaye.

Saare ke saare. $r = n$ par, $^{n} C_{n} = \frac{n!}{n! 0!} = 1$ : sab kuch lene ka ek hi tareeka hai.
Ek bhi nahi. Hum maante hain ki $^{n} C_{0} = 1$ : khaali selection hi zero objects ka akela selection hai. Formula bhi yahi kehta hai, kyunki $\frac{n!}{0! n!} = 1$ , isliye (6.5) $0 \leq r \leq n$ ke liye sahi hai.
Symmetry. $r$ ki jagah $n - r$ rakhne se denominator ke do factorials sirf aapas mein jagah badalte hain, isliye $^{n} C_{n - r} =^{n} C_{r}$ .
Barabar values. Agar $^{n} C_{a} =^{n} C_{b}$ , to ya to $a = b$ ya $a + b = n$ .

Symmetry

^{n} C_{n - r} =^{n} C_{r}

(6.6)

Symmetry ka ek seedha matlab hai, jo Fig. 6.12 mein dikhaya gaya hai: rakhne ke liye $r$ objects chunne ka har tareeka bahar rehne wale $n - r$ ko bhi tay kar deta hai, aur ulta bhi. 6 laddoos ke box se abhi khaane ke liye 2 chunna yeh tay kar deta hai ki baad ke liye kaunse 4 rakhoge, isliye $^{6} C_{2} =^{6} C_{4} = 15$ .

2/2… chhodne ke liye 4 chunne jaisa hi hai: $^{6} C_{2} =^{6} C_{4} = 15$ .

Fig. 6.12Kya lena hai, iski har choice yeh bhi choice hai ki kya chhodna hai.

Do lambi trays ek ke upar ek bani hain. Upar wali tray, label take 2, mein laddoos 2 aur 5 hain; neeche wali tray, label leave 4, mein laddoos 1, 3, 4 aur 6 hain. Pehle step mein take tray ka outline bold hai. Doosre step mein leave tray ka outline bold hota hai aur liye gaye do laddoos fade ho jaate hain, jo dikhata hai ki peeche chhoote chaar bhi ek selection banate hain.

Theorem 6.6Pascal's rule

$1 \leq r \leq n$ ke liye, $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ .

Kyunki $r! = r (r - 1)!$ aur $(n - r + 1)! = (n - r + 1) (n - r)!$ dono sahi hain, dono terms mein ek common factor hai. Short mein ise $K = \frac{n!}{(r - 1)! (n - r)!}$ kaho.

Tab $^{n} C_{r} = \frac{K}{r}$ aur $^{n} C_{r - 1} = \frac{K}{n - r + 1}$ , isliye unka sum $K$ times do chhote fractions ka sum hai. Common denominator lene par woh aaram se add ho jaate hain:

\begin{aligned} \frac{1}{r} + \frac{1}{n - r + 1} \\ = \frac{n + 1}{r (n - r + 1)} \end{aligned}

$K$ se multiply karo. $r (r - 1)! = r!$ aur $(n - r + 1) (n - r)! = (n - r + 1)!$ ek baar phir use karne par sum ban jaata hai

\begin{aligned} ^{n} C_{r} +^{n} C_{r - 1} \\ = \frac{(n + 1)!}{r! (n + 1 - r)!} \end{aligned}

aur (6.5) se right side $^{n + 1} C_{r}$ hai.

Ek counting argument bina algebra ke yahi rule deta hai. $n + 1$ logon mein se, jinmein ek Meera hai, $r$ chuno. Ya to Meera group mein hai, aur baaki $r - 1$ bache $n$ logon mein se $^{n} C_{r - 1}$ tareekon se aate hain, ya woh nahi hai, aur saare $r$ baaki $n$ mein se $^{n} C_{r}$ tareekon se aate hain. Cases overlap nahi karte, isliye add hote hain.

Numbers $^{n} C_{r}$ ko rows mein likho, har $n$ ke liye ek row, to Pascal's rule kehta hai ki har andar wali entry apne upar ki do entries ka sum hai (Fig. 6.13). Indian scholars ko yeh triangle Pascal se bahut pehle pata tha: commentator Halayudha ne 10th century mein ise Meru prastara ke roop mein describe kiya tha.

3/3Row 5 ki har andar wali entry isi tarah banti hai.

Fig. 6.13Har andar wali entry apne theek upar ki do entries ka sum hai.

Numbered boxes ki rows ek triangle banati hain, left mein label n equals 0 se n equals 5 tak: 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; aur ek halki aakhri row 1 5 10 10 5 1. Doosre step mein row 4 ke boxes 4 aur 6 highlight hote hain aur do arrows neeche row 5 ke box 10 ki taraf point karte hain, label 4C1 plus 4C2 equals 5C2 ke saath. Teesre step mein poori row 5 solid ho jaati hai.

Example 6.17NCERT Example 17 · p. 117

Symmetry use karna

Agar $^{n} C_{10} =^{n} C_{6}$ , to $n$ aur phir $^{n} C_{14}$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.18NCERT Example 18 · p. 117

Ek school panel

4 teachers aur 5 students mein se 4 members ka ek panel chunna hai. Kitne alag panels possible hain? Unmein se kitne mein theek 2 teachers aur 2 students hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.19NCERT Example 19 · p. 118

Pack se teen cards

52 cards ke ek standard pack se teen cards ek saath nikale jaate hain (upar wala box dekho). Possible hands ki number nikalo, aur unmein se kitne (i) saare ek suit ke hain; (ii) mein teen alag suits hain; (iii) saare face cards hain; (iv) mein do red cards aur ek black card hai; (v) saare ek colour ke hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Flash cards · 6

Arrange ya choose?

1 / 6

Practice · 11 questions

Practice 6.4: answers ke saath

Short answerEasyNCERT Exercise 6.4, Q1
Agar $^{n} C_{7} =^{n} C_{4}$ , to $^{n} C_{3}$ nikalo.
Answer dekho
Answer 165
$n = 7 + 4 = 11$ , isliye $^{11} C_{3} = \frac{11 \cdot 10 \cdot 9}{6} = 165.$
2
Short answerMediumNCERT Exercise 6.4, Q2
$n$ nikalo agar
1. $^{2 n} C_{2} :^{n} C_{2} = 22 : 5$
  Answer dekho
  Answer 6
  Ratio $\frac{2 (2 n - 1)}{n - 1}$ hai. Ise $\frac{22}{5}$ ke barabar rakhne par $20 n - 10 = 22 n - 22,$ milta hai, isliye $n = 6$ .
2. $^{2 n} C_{2} :^{n} C_{2} = 5 : 1$
  Answer dekho
  Answer 3
  $\frac{2 (2 n - 1)}{n - 1} = 5$ se $4 n - 2 = 5 n - 5$ milta hai, isliye $n = 3$ .
3
Short answerEasyNCERT Exercise 6.4, Q3
Ek circle par pandrah points hain.
1. Unhe pairs mein jodne wali kitni chords hain?
  Answer dekho
  Answer 105
  $^{15} C_{2} = 105$ .
2. Kitne triangles ke teeno corners in points mein se hain?
  Answer dekho
  Answer 455
  $^{15} C_{3} = 455$ .
Short answerEasyNCERT Exercise 6.4, Q4
6 boys aur 7 girls mein se 2 boys aur 4 girls ki ek team chunni hai. Aisi kitni teams possible hain?
Answer dekho
Answer 525
$^{6} C_{2} \times^{7} C_{4} = 15 \times 35 = 525.$
Short answerMediumNCERT Exercise 6.4, Q5
Ek basket mein 5 alag mangoes, 4 alag guavas aur 6 alag bananas hain. Har kind ke 2 lekar 6 fruits kitne tareekon se chun sakte ho?
Answer dekho
Answer 900
$^{5} C_{2} \times^{4} C_{2} \times^{6} C_{2} = 10 \times 6 \times 15 = 900.$
Short answerMediumNCERT Exercise 6.4, Q6
52 ke pack se kitne 5-card hands mein theek do aces hote hain?
Answer dekho
Answer 103776
4 aces mein se 2 aur baaki 48 cards mein se 3 chuno: $^{4} C_{2} \times^{48} C_{3} = 6 \times 17 296 = 103 776.$
Short answerMediumNCERT Exercise 6.4, Q7
Ek hockey squad mein 16 players hain, jinmein se 3 goalkeepers hain. 11 ki team mein theek ek goalkeeper hona chahiye. Kitni teams possible hain?
Answer dekho
Answer 858
$^{3} C_{1} \times^{13} C_{10} = 3 \times 286 = 858.$
Short answerEasyNCERT Exercise 6.4, Q8
Ek jar mein 7 green aur 4 yellow marbles hain, sab distinguishable. 3 green aur 2 yellow marbles kitne tareekon se nikale ja sakte hain?
Answer dekho
Answer 210
$^{7} C_{3} \times^{4} C_{2} = 35 \times 6 = 210.$
Short answerEasyNCERT Exercise 6.4, Q9
Ek student ko 12 mein se 6 electives chunne hain, aur 2 khaas electives compulsory hain. Choice kitne tareekon se ho sakti hai?
Answer dekho
Answer 210
Bache 10 mein se baaki 4 chuno: $^{10} C_{4} = 210$ .
Quick answerMedium
$^{10} C_{3} +^{10} C_{4}$ barabar hai
$^{20} C_{7}$ $^{10} C_{7} = 120$ $^{11} C_{4} = 330$ $^{11} C_{3} = 165$
Answer dekho
Answer (C) $^{11} C_{4} = 330$
$^{10} C_{4} +^{10} C_{3} =^{11} C_{4} = 330.$ Check karo: $120 + 210 = 330$ .
Quick answerMedium
Assertion (A): 10 books mein se trip par le jaane ke liye 3 chunne ke utne hi tareeke hain jitne ghar par chhodne ke liye 7 chunne ke. Reason (R): $^{n} C_{r} =^{n} C_{n - r}$ .
A aur R dono true hain, aur R, A ko explain karta haiA aur R dono true hain, lekin R, A ko explain nahi kartaA true hai, lekin R false haiA false hai, lekin R true hai
Answer dekho
Answer (A) A aur R dono true hain, aur R, A ko explain karta hai
Le jaane ke liye 3 ki har choice peeche chhoote 7 ko tay kar deti hai, isliye $^{10} C_{3} =^{10} C_{7} = 120$ . R, A ka general form hai.

Miscellaneous Examples

Baaki examples is chapter ke tools ko mila kar use karte hain: pehle select phir arrange, cases mein baantna, complement ginna, dictionary list mein aage badhna, aur gaps se objects ko alag rakhna.

Example 6.20NCERT Example 20 · p. 119

Pehle select, phir arrange

EDUCATION ke letters se, har ek ko zyada se zyada ek baar use karke, kitne 4-letter words (meaningful ho ya na ho) bana sakte ho jinmein theek 2 vowels aur 2 consonants hon?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.21NCERT Example 21 · p. 120

At least: cases se aur complement se

5 women aur 6 men mein se 4 ki ek committee chuni jaati hai. Woh kitne tareekon se ban sakti hai agar usmein (i) koi woman na ho; (ii) kam se kam ek woman aur kam se kam ek man ho; (iii) kam se kam 3 women hon?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.22NCERT Example 22 · p. 121

Dictionary list ka 50th word

PAPER ke letters ke saare arrangements dictionary order mein list kiye jaate hain. List mein kitne words hain, aur 50th word kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.23NCERT Example 23 · p. 121

Repeated digits se bade numbers

Digits 5, 0, 3, 5, 7, 0 ko, har ek ko utni hi baar jitna woh list mein aata hai, use karke 500 000 se bade kitne numbers ban sakte hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Kuch objects ko alag rakhne ke liye pehle baaki ko bithao aur unke beech ke gaps use karo (Fig. 6.14).

3/33 girls ko 3 alag gaps mein order se rakho: $^{5} P_{3} = 60$ tareeke. Koi do girls padosi nahi hain.

Fig. 6.14Alag gaps mein rakhi girls kabhi ek doosre ke bagal mein nahi baith sakti.NCERT Example 24 · p. 122

B likhe chaar square tiles ek row mein khade hain aur unke aas-paas paanch dashed gap boxes hain, har end par ek aur beech mein teen; text kehta hai boys first: 4! orders. Doosre step mein gaps ka outline bold hota hai aur unpar cross lagte hain, G likhe teen round tiles upar wait karte hain, aur text kehta hai 5 gaps for the girls. Teesre step mein teeno girls pehle, teesre aur chauthe gap mein aa jaati hain, aur text kehta hai girls in 3 of the 5 gaps.

Example 6.24NCERT Example 24 · p. 122

Koi do girls bagal mein nahi

4 boys aur 3 girls 7 chairs ki row mein kitne tareekon se baith sakte hain taaki koi do girls bagal mein na baithein?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

TIGER ke letters se 1 vowel aur 2 consonants wale kitne 3-letter words ban sakte hain?

Answer dekho

Answer 36

Vowels I, E; consonants T, G, R.

^{2} C_{1} \times^{3} C_{2} = 6

letter sets chuno, har ek

3! = 6

tareekon se arrange hota hai: 36.

Practice · 14 questions

Practice questions answers ke saath (NCERT “Chapter 6 par Miscellaneous Exercise”)

Short answerMediumNCERT Miscellaneous Exercise, Q1
FRACTION ke letters ko, har ek ko zyada se zyada ek baar use karke, kitne 4-letter words (meaningful ho ya na ho) mein theek 1 vowel aur 3 consonants hote hain?
Answer dekho
Answer 720
Vowels hain A, I, O; consonants hain F, R, C, T, N. $^{3} C_{1} \times^{5} C_{3} \times 4! = 3 \times 10 \times 24 = 720.$
Short answerMediumNCERT Miscellaneous Exercise, Q2
COUNTRIES ke saare letters ke kitne arrangements mein vowels saath aur consonants saath rehte hain?
Answer dekho
Answer 5760
Vowels O, U, I, E ek block banate hain (andar $4!$ orders) aur consonants C, N, T, R, S doosra ( $5!$ orders). Blocks $2!$ orders mein aate hain: $2 \times 24 \times 120 = 5760.$
3
Short answerMediumNCERT Miscellaneous Exercise, Q3
7 men aur 5 women mein se 6 ki ek committee banani hai. Woh possible committees gino jinmein
1. theek 2 women hon?
  Answer dekho
  Answer 350
  $^{5} C_{2} \times^{7} C_{4} = 10 \times 35 = 350.$
2. kam se kam 3 women hon?
  Answer dekho
  Answer 462
  3, 4 ya 5 women: $10 \cdot 35 + 5 \cdot 21 + 1 \cdot 7 = 462.$
3. zyada se zyada 1 woman ho?
  Answer dekho
  Answer 112
  0 ya 1 woman: $^{7} C_{6} +^{5} C_{1} \times^{7} C_{5} = 7 + 105 = 112.$
Short answerHardNCERT Miscellaneous Exercise, Q4
BALLOON ke letters ke saare arrangements dictionary order mein list kiye jaate hain. L se shuru hone wale pehle word se pehle kitne words aate hain?
Answer dekho
Answer 360
A se shuru karne par letters B, L, L, O, O, N se $\frac{6!}{2! 2!} = 180$ words milte hain. B se shuru karne par letters A, L, L, O, O, N se aur 180. To 360 words pehle aate hain.
Short answerHardNCERT Miscellaneous Exercise, Q5
Digits 0, 2, 3, 5, 6, 8 se, koi digit repeat kiye bina, 5 se divisible kitne 5-digit numbers ban sakte hain?
Answer dekho
Answer 216
Aakhri digit 0: baaki 5 digits mein se 4 ko aage arrange karo, $^{5} P_{4} = 120$ . Aakhri digit 5: pehla digit 2, 3, 6 ya 8 (4 tareeke), phir beech ke liye $^{4} P_{3} = 24$ , to 96. Total $120 + 96 = 216$ .
Short answerMediumNCERT Miscellaneous Exercise, Q6
English alphabet ke 26 letters mein se 5 vowels hain. Woh 4-letter strings gino jo ek vowel aur teen alag consonants use karti hain.
Answer dekho
Answer 159600
$^{5} C_{1} \times^{21} C_{3} \times 4! = 5 \times 1330 \times 24 = 159 600.$
Short answerMediumNCERT Miscellaneous Exercise, Q7
Ek test mein Part A aur Part B hain, dono mein 6-6 questions. Candidate ko 7 questions answer karne hain, har part se kam se kam 2. Woh 7 questions ke kitne alag sets chun sakti hai?
Answer dekho
Answer 780
(A, B) ke possible splits (2, 5), (3, 4), (4, 3), (5, 2) hain: $15 \cdot 6 + 20 \cdot 15 + 15 \cdot 20 + 6 \cdot 15 = 780.$
Short answerMediumNCERT Miscellaneous Exercise, Q8
52 ke pack se kitne 4-card hands mein theek ek heart hota hai?
Answer dekho
Answer 118807
$^{13} C_{1} \times^{39} C_{3} = 13 \times 9139 = 118 807.$
Short answerHardNCERT Miscellaneous Exercise, Q9
Chaar boys aur 3 girls 1 se 7 tak numbered 7 chairs ki row mein baithte hain. Girls sirf odd-numbered chairs par baith sakti hain. Kitne seatings possible hain?
Answer dekho
Answer 576
Odd chairs 1, 3, 5, 7 hain. Girls in 4 mein se 3 chairs order se leti hain: $^{4} P_{3} = 24$ . Boys bachi 4 chairs bharte hain: $4! = 24$ . Total $24 \times 24 = 576$ .
Short answerHardNCERT Miscellaneous Exercise, Q10
20 ki class mein se 8 students ko field trip par jaana hai. Unmein se do twins hain jo sirf saath hi jaayenge. Kitne groups possible hain?
Answer dekho
Answer 62322
Twins jaate hain: baaki 18 mein se 6 aur chuno, $^{18} C_{6} = 18 564$ . Twins rukte hain: 18 mein se saare 8 chuno, $^{18} C_{8} = 43 758$ . Total $62 322$ .
Short answerMediumNCERT Miscellaneous Exercise, Q11
POSSESSIONS ke letters ke kitne arrangements mein saare S saath hain?
Answer dekho
Answer 2520
POSSESSIONS mein 11 letters hain: S paanch baar, O do baar, aur P, E, I, N. S block P, O, O, E, I, N ke saath 7 units banata hai jismein O do baar hai: $\frac{7!}{2!} = 2520$ .
Quick answerMedium
Decagon (10 sides wala polygon) ke diagonals ki number hai
45904035
Answer dekho
Answer (D) 35
$^{10} C_{2} = 45$ segments vertices ke pairs ko jodte hain; unmein se 10 sides hain, to 35 diagonals bachte hain.
Quick answerHard
$8 \times 8$ chessboard par har size ke kitne rectangles mil sakte hain?
Answer dekho
Answer 1296
Ek rectangle 9 horizontal lines mein se 2 aur 9 vertical lines mein se 2 chunne se tay hota hai: $^{9} C_{2} \times^{9} C_{2} = 36 \times 36 = 1296.$
Quick answerHard
Assertion (A): $^{10} C_{4} =^{10} C_{6}$ . Reason (R): Har $r$ ke liye $^{n} P_{r} =^{n} P_{n - r}$ .
A aur R dono true hain, aur R, A ko explain karta haiA aur R dono true hain, lekin R, A ko explain nahi kartaA true hai, lekin R false haiA false hai, lekin R true hai
Answer dekho
Answer (C) A true hai, lekin R false hai
Symmetry se A true hai (dono 210 ke barabar). R false hai: $^{10} P_{4} = 5040$ lekin $^{10} P_{6} = 151 200$ . Order ki wajah se dono counts alag hote hain.

Recap

Summary aur key points

Fundamental principle of counting: ek ke baad ek hone wale tasks, jo $m_{1}, m_{2}, \dots, m_{k}$ tareekon se hote hain, saath milkar $m_{1} m_{2} \dots m_{k}$ tareekon se ho sakte hain (Section 6.2).
Successive stages multiply hoti hain, disjoint cases add hote hain, aur sabse restricted place pehle bhari jaati hai (Example 6.3).
Permutations orderings ginte hain; $^{n} P_{r} = n (n - 1) \dots (n - r + 1)$ (Theorem 6.1).
$n! = 1 \times 2 \times \dots \times n,$ $n! = n (n - 1)!$ aur $0! = 1$ (Section 6.3.2).
$0 \leq r \leq n$ ke liye $^{n} P_{r} = \frac{n!}{(n - r)!}$ , isliye $^{n} P_{n} = n!$ aur $^{n} P_{0} = 1$ ((6.3)).
Repetition allowed ho to $n$ objects se $r$ places bharne par $n^{r}$ arrangements milte hain (Theorem 6.2).
Har kind ke $p_{1}, p_{2}, \dots, p_{k}$ alike objects hon to $n$ objects ke $\frac{n!}{p_{1}! p_{2}! \dots p_{k}!}$ arrangements hote hain (Theorem 6.4).
Saath: ek block glue karo aur uske andar ke orders se multiply karo; saare saath nahi: total minus saath wale (Example 6.14).
Combination ek aisa selection hai jismein order matter nahi karta, aur $^{n} P_{r} =^{n} C_{r} \times r!$ (Theorem 6.5).
$^{n} C_{r} = \frac{n!}{r! (n - r)!}$ , saath mein $^{n} C_{0} =^{n} C_{n} = 1$ aur $^{n} C_{n - r} =^{n} C_{r}$ ((6.5)).
Pascal's rule kehta hai: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ (Theorem 6.6).
Mixed problems: pehle select phir arrange, at least ko cases mein baanto ya complement use karo, aur gaps se objects ko alag rakho (Miscellaneous Examples).

Timeline

Historical note: alag-alag cultures mein counting

Ancient India
Sushruta ke 63 tastes
Medical text Sushruta Samhita, jise paramparik roop se lagbhag 6th century BCE ka maana jaata hai, kehta hai ki 6 basic tastes ek, do ya zyada ek saath lene par 63 mixtures dete hain. Check: $^{6} C_{1} +^{6} C_{2} + \dots +^{6} C_{6} = 63.$
lagbhag 3rd century BCE
Pingala metres ginte hain
Pingala ke Chhandas Shastra ne verse ki ek line mein long aur short syllables ke patterns list karne aur ginne ke rules diye.
Shuruaati Jain texts
Vikalpa
Jain scholars ne selections aur arrangements ki counting ko, jise woh vikalpa kehte the, apne aap mein ek alag subject maana.
Ancient China
I Ching ke hexagrams
I Ching (Book of Changes) chhe lines wale figures use karta hai, har line poori ya tooti hui: kul $2^{6} = 64$ hexagrams.
lagbhag 850
Mahavira ka rule
Mahavira ke Ganita Sara Sangraha ne combinations ki number ke liye ek general rule diya, jo $^{n} C_{r}$ ke formula ke barabar hai.
10th century
Meru prastara
Halayudha ne Pingala par commentary likhte hue numbers ke us triangle ko describe kiya jise aaj Pascal ke naam se jaana jaata hai.
lagbhag 1140
Abraham ibn Ezra
Unhone us samay pata saat planets ke possible conjunctions gine, do, teen ya zyada ek saath lekar: kul 120.
1150
Bhaskara II ki Lilavati
Anka Pasha (numbers ka jaal) naam ka ek chapter digits ke arrangements nikalta hai, jismein repeat hone wale digits bhi shaamil hain.
1321
Levi ben Gerson
Unki Maaseh Hoshev ne mathematical induction ke ek shuruaati form se permutations aur combinations ke formulas prove kiye.
1713
Bernoulli ki Ars Conjectandi
Jacob Bernoulli ki kitaab, jo unki death ke baad publish hui, ne permutations aur combinations ka pehla poora treatment diya.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 6.1, Q1	Practice 6.1 · Q1
NCERT Exercise 6.1, Q2	Practice 6.1 · Q2
NCERT Exercise 6.1, Q3	Practice 6.1 · Q3
NCERT Exercise 6.1, Q4	Practice 6.1 · Q4
NCERT Exercise 6.1, Q5	Practice 6.1 · Q5
NCERT Exercise 6.1, Q6	Practice 6.1 · Q6
NCERT Exercise 6.2, Q1	Practice 6.2 · Q1
NCERT Exercise 6.2, Q2	Practice 6.2 · Q2
NCERT Exercise 6.2, Q3	Practice 6.2 · Q3
NCERT Exercise 6.2, Q4	Practice 6.2 · Q4
NCERT Exercise 6.2, Q5	Practice 6.2 · Q5
NCERT Exercise 6.3, Q1	Practice 6.3 · Q1
NCERT Exercise 6.3, Q2	Practice 6.3 · Q2
NCERT Exercise 6.3, Q3	Practice 6.3 · Q3
NCERT Exercise 6.3, Q4	Practice 6.3 · Q4
NCERT Exercise 6.3, Q5	Practice 6.3 · Q5
NCERT Exercise 6.3, Q6	Practice 6.3 · Q6
NCERT Exercise 6.3, Q7	Practice 6.3 · Q7
NCERT Exercise 6.3, Q8	Practice 6.3 · Q8
NCERT Exercise 6.3, Q9	Practice 6.3 · Q9
NCERT Exercise 6.3, Q10	Practice 6.3 · Q10
NCERT Exercise 6.3, Q11	Practice 6.3 · Q11
NCERT Exercise 6.4, Q1	Practice 6.4 · Q1
NCERT Exercise 6.4, Q2	Practice 6.4 · Q2
NCERT Exercise 6.4, Q3	Practice 6.4 · Q3
NCERT Exercise 6.4, Q4	Practice 6.4 · Q4
NCERT Exercise 6.4, Q5	Practice 6.4 · Q5
NCERT Exercise 6.4, Q6	Practice 6.4 · Q6
NCERT Exercise 6.4, Q7	Practice 6.4 · Q7
NCERT Exercise 6.4, Q8	Practice 6.4 · Q8
NCERT Exercise 6.4, Q9	Practice 6.4 · Q9
NCERT Miscellaneous Exercise, Q1	Chapter 6 par Miscellaneous Exercise · Q1
NCERT Miscellaneous Exercise, Q2	Chapter 6 par Miscellaneous Exercise · Q2
NCERT Miscellaneous Exercise, Q3	Chapter 6 par Miscellaneous Exercise · Q3
NCERT Miscellaneous Exercise, Q4	Chapter 6 par Miscellaneous Exercise · Q4
NCERT Miscellaneous Exercise, Q5	Chapter 6 par Miscellaneous Exercise · Q5
NCERT Miscellaneous Exercise, Q6	Chapter 6 par Miscellaneous Exercise · Q6
NCERT Miscellaneous Exercise, Q7	Chapter 6 par Miscellaneous Exercise · Q7
NCERT Miscellaneous Exercise, Q8	Chapter 6 par Miscellaneous Exercise · Q8
NCERT Miscellaneous Exercise, Q9	Chapter 6 par Miscellaneous Exercise · Q9
NCERT Miscellaneous Exercise, Q10	Chapter 6 par Miscellaneous Exercise · Q10
NCERT Miscellaneous Exercise, Q11	Chapter 6 par Miscellaneous Exercise · Q11

Quick check: MCQs

6.2 Fundamental Principle of Counting4 questions
6.3.2 Factorial notation4 questions
6.3.3 nPr ke formula ka derivation2 questions
6.3.4 Jab saare objects distinct na hon tab permutations4 questions
6.4 Combinations4 questions
Miscellaneous Examples3 questions

Glossary · 7

Key terms (English + Hindi)

addition principle: Agar outcomes aise cases mein bantte hain jo saath nahi ho sakte, to total case counts ka sum hota hai.Chapter mein pehli baar yahan aaya →

combinationसंचयsanchay: Objects ka aisa selection jismein order matter nahi karta; $^{n} C_{r}$ , $n$ alag objects mein se $r$ wale aise selections ginta hai.Chapter mein pehli baar yahan aaya →

factorialक्रमगुणितkramgunit: Natural number $n$ ke liye $n! = 1 \times 2 \times \dots \times n$ , aur $0! = 1$ .
fundamental principle of countingगणना का आधारभूत सिद्धांतganana ka aadharbhoot siddhant: Agar ek task $m$ tareekon se aur phir doosra $n$ tareekon se ho sakta hai, to dono order mein $m n$ tareekon se ho sakte hain.Chapter mein pehli baar yahan aaya →

Pascal's rule: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ : Pascal's triangle ki har andar wali entry apne upar ki do entries ka sum hai.
permutationक्रमचयkramchay: Objects ke set mein se kuch ya saare objects ka ek definite order mein arrangement; $^{n} P_{r}$ , $n$ alag objects mein se $r$ wale aise arrangements ginta hai.Chapter mein pehli baar yahan aaya →

tree diagramवृक्ष आरेखvriksh aarekh: Successive choices ki branching picture jismein start se kisi leaf tak har path ek outcome hai.Chapter mein pehli baar yahan aaya →

6.1 Introduction

6.2 Fundamental Principle of Counting

6.3 Permutations

6.3.1 Jab saare objects distinct hon tab permutations

6.3.2 Factorial notation

6.3.3 nPr ke formula ka derivation

6.3.4 Jab saare objects distinct na hon tab permutations

6.4 Combinations

Miscellaneous Examples

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye