Permutations and Combinations: Class 11 Maths · Chapter 6

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

90 min
1 simulation
50 practice questions
21 MCQs

What you will learn in Permutations and Combinations

Apply the fundamental principle of counting, and the addition principle for separate cases, to count outcomes without listing them
Use factorial notation, including 0! = 1, to simplify expressions and solve equations
Derive and use nPr = n!/(n − r)! and n^r to count arrangements of distinct objects without and with repetition
Count arrangements when some objects are alike, and handle fixed places, blocks kept together and objects kept apart
Derive nCr = n!/(r!(n − r)!) from nPr = nCr × r!, and use the symmetry property and Pascal's rule
Decide whether a problem needs arrangements or selections, and solve mixed problems at board and entrance level

Original companion text aligned to NCERT Mathematics Class 11, Chapter 6, pp. 100–125. Not affiliated with or endorsed by NCERT.

6.1 Introduction

Look at the number plate of any car or scooter in India. After the state code and the district number comes a short series of letters and then four digits, as in MH 12 KX 4071. Suppose a district may fill that tail with any two letters followed by any four digits. How many different tails can it issue before it runs out?

Listing them would take weeks; the answer is $26 \times 26 \times 10^{4}$ , or 67.6 lakh. This chapter reaches such numbers by reasoning instead. Two questions run through it: in how many ways can objects be arranged in order, and in how many ways can a group be selected when order does not matter? The first leads to permutations, the second to combinations. Both rest on one simple rule, the subject of the next section.

6.2 Fundamental Principle of Counting

A school canteen makes two kinds of paratha, aloo and gobhi, and serves three drinks: lassi, chai and chaas. A meal is one paratha with one drink. Whichever paratha you take, all three drinks are still on offer, so each paratha starts 3 meals and the two parathas give $2 \times 3 = 6$ meals. A tree diagram makes this visible (Fig. 6.1): every path from the top to a leaf is one meal.

3/3Six leaves, one for each meal: $2 \times 3 = 6$ .

Fig. 6.1Each of the 2 first choices opens the same number of second choices, 3, so there are 6 paths.NCERT Fig. 6.1 · p. 101

A tree drawn from the top down. In the first step a single start point splits into two boxes, aloo and gobhi, for the choice of paratha. In the second step each paratha box splits into three dashed boxes, lassi, chai and chaas, so six leaves appear along the bottom. In the third step the six leaves are numbered 1 to 6 and the line 2 times 3 equals 6 meals appears underneath.

Now add a stage. A wedding thali has one sabzi out of 3, one dal out of 2 and one sweet out of 4. Each of the $3 \times 2 = 6$ sabzi–dal pairs can be finished with any of the 4 sweets, so there are $6 \times 4 = 24$ different thalis. Drawing that tree needs 24 leaves; multiplying needs one line. This rule is the fundamental principle of counting; many books call it the multiplication principle.

PrincipleFundamental principle of counting

Count a two-stage process stage by stage. If stage one has $m$ possible outcomes, and for each of them stage two has $n$ possible outcomes, then the whole process has $m \times n$ possible outcomes. With more stages in a fixed order, keep multiplying.

Multiplication principle

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

(6.1)

where

$m_{i}$: the number of ways to do task $i$ , once tasks 1 to $i - 1$ are done
$N$: the number of ways to do all $k$ tasks in order

Example 6.1NCERT Example 1 · p. 102

Five letters in a row

How many five-letter strings (they need not be words) can be made from the letters of PLANT if no letter may be used twice? How many can be made if letters may repeat?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.2NCERT Example 2 · p. 103

Two rings of a rangoli

A rangoli has an inner ring and an outer ring. Each ring is filled with one colour chosen from 5 colours, and the two rings must have different colours. How many colourings are possible?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.3NCERT Example 3 · p. 103

Odd numbers, digits may repeat

How many three-digit odd numbers can be made from the digits 2, 3, 4, 5, 6, 7 if digits may repeat?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.4NCERT Example 4 · p. 103

Codes of different lengths

A club's door code is a row of 2, 3 or 4 different letters taken from A, B, C, D, E, F. How many codes are possible?

Try it yourself first. Then reveal the solution one step at a time.

Counting is also the first step of probability, which you meet in Chapter 14. Two dice thrown together give $6 \times 6 = 36$ ordered outcomes, because the first die's 6 faces each pair with the second die's 6. Only 6 of these outcomes are doubles, so a double turns up about one throw in six. The lab below lets you test that count.

SimulationTwo dice: 6 × 6 outcomes

A two-dice lab. Buttons roll the pair once, 10, 100 or 1000 times. A bar chart shows how often each total from 2 to 12 has appeared next to its theoretical share, and a readout compares the fraction of doubles with the theoretical value of one sixth. Nothing moves until the learner presses a roll button.

Fig. 6.2Each throw is one of

6 \times 6 = 36

ordered pairs, and 6 of them are doubles. Press a roll button to compare the counts with the theory.

Try this

Before rolling: how many ordered outcomes do two dice have, and how many of them are doubles?
Roll 1000 times. Is the share of doubles close to $\frac{6}{36} = \frac{1}{6}$ ?
Why does a total of 7 appear more often than a total of 2? Count the ordered pairs that give each.

Practice · 7 questions

Practice 6.1: questions with answers

1
Short answerEasyNCERT Exercise 6.1, Q1
Only the digits 2, 4, 5, 7, 8 and 9 are allowed. Count the 4-digit numbers you can write if
1. digits may repeat?
  Show answer
  Answer 1296
  Each place has 6 choices: $6^{4} = 1296$ .
2. no digit may repeat?
  Show answer
  Answer 360
  $6 \times 5 \times 4 \times 3 = 360$ .
Short answerMediumNCERT Exercise 6.1, Q2
Digits may repeat. Using 1, 2, 3, 4, 5, 6 and 7 only, how many 4-digit multiples of 5 can you write?
Show answer
Answer 343
With no 0 available, the units digit must be 5: 1 way. The other three places take any of 7 digits: $7^{3} = 343$ .
Short answerEasyNCERT Exercise 6.1, Q3
How many 3-letter codes can be made from the 12 letters O to Z, with no letter used twice?
Show answer
Answer 1320
$12 \times 11 \times 10 = 1320.$
Short answerMediumNCERT Exercise 6.1, Q4
A 6-digit locker code must begin with 40, and no digit may appear twice. How many codes are possible?
Show answer
Answer 1680
Eight digits remain for four places: $8 \times 7 \times 6 \times 5 = 1680.$
Short answerEasyNCERT Exercise 6.1, Q5
A die is rolled once and then a coin is tossed twice. How many different records of the three results are possible?
Show answer
Answer 24
$6 \times 2 \times 2 = 24$ .
Short answerEasyNCERT Exercise 6.1, Q6
Seven lanterns of different colours are available. A festival signal hangs 3 of them one below another. How many signals are possible?
Show answer
Answer 210
$7 \times 6 \times 5 = 210$ . Order matters because the top-to-bottom order is part of the signal.
Quick answerHard
How many 3-digit numbers have at least one repeated digit?
252648900271
Show answer
Answer (A) 252
There are 900 three-digit numbers. Those with all digits different number $9 \times 9 \times 8 = 648$ : the first digit cannot be 0, the second can. So $900 - 648 = 252$ have a repeat.

6.3 Permutations

Swap two books on a shelf and the shelf looks different, although the same books are there. Whenever the order of the objects is part of what we count, each possible order is called a permutation (क्रमचय). The strings of Example 6.1 were permutations of the letters of PLANT.

Definition

Permutation

An ordering of objects: some or all of them placed one after another, where a change of order gives a different permutation.

ExampleFrom the letters of PLANET, LANE and ELAN are two different permutations of 4 letters.

A permutation may use only some of the objects. A 4-letter code built from the letters of PLANET, with no letter repeated, is a permutation of 6 letters taken 4 at a time, and the slot method gives $6 \times 5 \times 4 \times 3 = 360$ such codes. The next subsections turn such products into formulas.

6.3.1 Permutations when all the objects are distinct

Take $n$ different objects and $r$ empty places in a row, where $0 < r \leq n$ . Fill the places one at a time and never reuse an object. Fig. 6.3 does this with six letter tiles and three places.

4/44 tiles are left, so there are $6 \times 5 \times 4 = 120$ arrangements.

Fig. 6.3Every tile that is used leaves one fewer for the next place.

Three dashed boxes sit at the top and six letter tiles, A to F, lie in a row below. In the second step tile C moves up into the first box and the label 6 ways appears under it. In the third step tile F moves into the second box, labelled 5 ways, because only five tiles were left. In the fourth step tile A fills the third box, labelled 4 ways, and the product 6 times 5 times 4 equals 120 appears at the bottom. Dashed outlines mark where the used tiles were.

Theorem 6.1Arrangements of distinct objects

Let $0 < r \leq n$ . If no object may be used twice, $r$ objects taken from $n$ different ones can be lined up in $n (n - 1) (n - 2) \dots (n - r + 1)$ ways. This number is written $^{n} P_{r}$ .

The product has $r$ factors, which is clumsy when $r$ is large. A new notation fixes that.

6.3.2 Factorial notation

For a natural number $n$ , the product of all the natural numbers from 1 up to $n$ is written $n!$ and read $n$ factorial (क्रमगुणित). So $1! = 1$ , $2! = 2$ , $3! = 6$ , $4! = 24$ and $5! = 120$ . Factorials grow very fast: $10!$ is already more than 36 lakh.

Definition

Factorial

For a natural number $n$ , $n! = 1 \times 2 \times 3 \times \dots \times n .$ By agreement, $0! = 1$ .

Example $6! = 720$

Peeling off the largest factor leaves a smaller factorial: $6! = 6 \times 5!$ and $5! = 5 \times 4!$ . Repeating the step gives $n! = n (n - 1) (n - 2)!$ for $n \geq 2$ , and so on. The basic step is:

Factorial, one step down

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

(6.2)

Why should $0!$ be 1? Read (6.2) backwards: each step down divides by the number you leave behind. Fig. 6.4 walks down from $4!$ . The last step, from $1!$ to $0!$ , divides 1 by 1 and leaves 1. Agreeing that $0! = 1$ keeps the pattern unbroken, and it will make several formulas work at their edge cases.

3/3Divide by 1: the pattern says $0! = 1$ .

Fig. 6.4Going down one step divides by the number you leave behind.

A staircase of boxes descends from upper left to lower right. In the first step three boxes show 4! equals 24, 3! equals 6 and 2! equals 2, joined by curved arrows labelled divide by 4 and divide by 3. In the second step a fourth box appears, 1! equals 1, reached by dividing by 2. In the third step a highlighted last box appears, 0! equals 1, reached by dividing 1 by 1.

Example 6.5NCERT Example 5 · p. 106

Evaluating factorials

Evaluate (i) $6!$ (ii) $8!$ (iii) $8! - 6!$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 6.6NCERT Example 6 · p. 106

Cancelling factorials

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

Try it yourself first. Then reveal the solution one step at a time.

Example 6.7NCERT Example 7 · p. 106

Substituting into a factorial expression

Evaluate $\frac{n!}{r! (n - r)!}$ when $n = 7$ and $r = 3$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 6.8NCERT Example 8 · p. 106

An unknown in a factorial equation

Find $x$ if $\frac{1}{7!} + \frac{1}{8!} = \frac{x}{9!}$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Which is equal to

\frac{10!}{8!}

\frac{10}{8}

90

2!

80

Show answer

Answer (B) $90$

10! = 10 \times 9 \times 8!

, so the quotient is

10 \times 9 = 90

. Dividing 10 by 8 ignores what the factorial sign means.

Practice · 5 questions

Practice 6.2: questions with answers

1
Short answerEasyNCERT Exercise 6.2, Q1
Evaluate
1. $7!$
  Show answer
  Answer 5040
  $7! = 720 \times 7 = 5040$ .
2. $5! - 4!$
  Show answer
  Answer 96
  $4! (5 - 1) = 24 \times 4 = 96.$
Quick answerEasyNCERT Exercise 6.2, Q2
True or false: $(2 \times 3)! = 2! \times 3!$ .
TrueFalse
Show answer
Answer False
$(2 \times 3)! = 6! = 720$ , while $2! \times 3! = 2 \times 6 = 12$ .
Short answerEasyNCERT Exercise 6.2, Q3
Compute $\frac{10!}{7! 3!}$ .
Show answer
Answer 120
$\frac{10 \times 9 \times 8}{6} = 120$ .
Short answerMediumNCERT Exercise 6.2, Q4
Find $x$ if $\frac{1}{4!} - \frac{1}{5!} = \frac{x}{6!}$ .
Show answer
Answer 24
$\frac{6!}{4!} - \frac{6!}{5!} = 30 - 6 = 24.$
5
Short answerEasyNCERT Exercise 6.2, Q5
Evaluate $\frac{n!}{(n - r)!}$ when
1. $n = 8$ , $r = 3$
  Show answer
  Answer 336
  $8 \times 7 \times 6 = 336$ .
2. $n = 10$ , $r = 4$
  Show answer
  Answer 5040
  $10 \times 9 \times 8 \times 7 = 5040.$

6.3.3 Derivation of the formula for nPr

DerivationFrom a long product to factorials

Start from Theorem 6.1: $^{n} P_{r}$ is the product of the $r$ numbers from $n$ down to $n - r + 1$ . The numbers below them, from $n - r$ down to 1, are missing. Together they make $(n - r)!$ .

Multiply and divide by this missing tail. The numerator becomes the complete product $n!$ , so for $0 \leq r \leq n$ :

Permutations of distinct objects

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

(6.3)

Fig. 6.5 shows the step with $n = 7$ and $r = 3$ .

3/3The extra factors cancel, so $^{7} P_{3} = \frac{7!}{4!} = 210$ .

Fig. 6.5Multiplying and dividing by the missing tail turns the product into a quotient of factorials.

In the first step three solid boxes hold 7, 6 and 5, the three factors of 7P3. In the second step four dashed boxes, 4, 3, 2 and 1, join the row to make 7!, a fraction bar appears, and the same four dashed boxes appear under the bar as 4!. In the third step slanting lines strike out the dashed boxes above and below the bar, which fade, and the result 7 times 6 times 5 equals 210 is shown.

Two edge cases come free. With $r = n$ , $^{n} P_{n} = \frac{n!}{0!} = n!$ : all $n$ objects in a row. With $r = 0$ the formula gives $\frac{n!}{n!} = 1$ , which fits: the empty row is the only row with no objects in it. So (6.3) holds for every $r$ from 0 to $n$ .

Theorem 6.2Arrangements with repetition

When objects may be reused as often as we like, $r$ places can be filled from $n$ different objects in $n^{r}$ ways.

The formulas redo our slot counts in one line: $^{5} P_{5} = 5! = 120$ and $5^{5} = 3125$ for the PLANT strings of Example 6.1, and $^{6} P_{4} = \frac{6!}{2!} = 360$ for the PLANET codes. They settle new cases just as fast. If 10 passengers are waiting for 3 different berths, lower, middle and upper, the berths can be given out in $^{10} P_{3} = 720$ ways; order matters because the berths differ.

Quick check

1 of 2

Which expression equals

^{9} P_{4}

\frac{9!}{4!}

9^{4}

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

\frac{9!}{5!}

Show answer

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

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

Dividing by

4!

uses

r

where

n - r

belongs.

6.3.4 Permutations when all the objects are not distinct

So far every object was different. What changes when some are alike? Take the word MAA, Hindi for mother, with its two identical A's. Tag them for a moment as $A_{1}$ and $A_{2}$ , so that all three letters differ; then there are $3! = 6$ arrangements. Now remove the tags. $M A_{1} A_{2}$ and $M A_{2} A_{1}$ both become MAA, and each untagged word comes from exactly $2! = 2$ tagged ones (Fig. 6.6). So MAA has $\frac{6}{2} = 3$ arrangements.

3/3Each pair becomes one word: $\frac{3!}{2!} = 3$ words.

Fig. 6.6Removing the tags merges orders that differ only by swapping the identical letters.

A column of six boxes lists the tagged orders M A1 A2, M A2 A1, A1 M A2, A2 M A1, A1 A2 M and A2 A1 M under the heading tagged: 3! equals 6. In the second step curly braces join the boxes in pairs that differ only by swapping A1 and A2. In the third step each pair points to one untagged word on the right, MAA, AMA and AAM, under the heading tags removed, with 6 divided by 2! equals 3 below.

The same reasoning scales up. BANANA has 6 letters: three A's, two N's and one B. Tagging gives $6!$ orders. One untagged word is produced by $3!$ ways of shuffling the A tags and, independently, $2!$ ways of shuffling the N tags, so by $3! \times 2!$ tagged orders. Hence BANANA has $\frac{6!}{3! 2!} = 60$ arrangements.

Theorem 6.3One kind repeated

Suppose $p$ of $n$ objects are identical and the other $n - p$ differ from them and from each other. Then all $n$ can be put in a row in $\frac{n!}{p!}$ distinguishable ways.

Theorem 6.4Several kinds repeated

Sort $n$ objects into kinds: $p_{1}$ identical objects of one kind, $p_{2}$ of another, and so on up to $p_{k}$ of a $k$ th kind, with any objects left over all different. Rows that use all $n$ objects can then be made in $\frac{n!}{p_{1}! p_{2}! \dots p_{k}!}$ distinguishable ways.

InteractiveWord arrangement lab

A word lab. Preset buttons load MAA, BANANA, PAPER and TIRUPATI, and a text box accepts any word of up to 12 letters. Letter tiles show the word, with repeated letters dashed and tagged 1, 2, 3. A status panel lists the letter counts, the number of arrangements as n factorial divided by the factorials of the repeat counts, and the rank of the word in dictionary order. A number box with a Find it button shows the word at any position in the dictionary list. When there are at most 360 arrangements, all of them are listed in dictionary order with the typed word outlined. Nothing moves until the learner acts.

Fig. 6.7Type a word, or tap a preset, to see its letter counts, the number of arrangements and where the word sits in dictionary order.

Example 6.9NCERT Example 9 · p. 110

Letters of a temple town

How many arrangements are there of all the letters of TIRUPATI?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.10NCERT Example 10 · p. 110

Five-digit numbers without repeats

Using each of the digits 1 to 8 at most once, how many 5-digit numbers can you make?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.11NCERT Example 11 · p. 110

When 0 cannot lead

Using each of the seven digits 0 to 6 at most once, how many 4-digit numbers can you make?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.12NCERT Example 12 · p. 111

Solving for n

Find $n$ if (i) $^{n} P_{4} = 20^{n} P_{2}$ ; (ii) $^{n} P_{3} :^{n - 1} P_{3} = 4 : 3$ . In both, $n \geq 4$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 6.13NCERT Example 13 · p. 112

Solving for r, and rejecting a root

Find $r$ if $7 \cdot^{6} P_{r} = 30 \cdot^{7} P_{r - 1}$ .

Try it yourself first. Then reveal the solution one step at a time.

Conditions such as these letters must stay together are handled by gluing: treat the group as one block, arrange the blocks, then arrange the letters inside the block. For not all together, count every arrangement and subtract the glued ones. Fig. 6.8 shows the gluing step.

3/3Arrange the 7 units in $7!$ ways and the vowels inside the block in $3!$ ways.

Fig. 6.8Glue the group into one block, arrange the units, then arrange inside the block.

A row of nine letter tiles spells TRIANGLES, with the vowel tiles I, A and E drawn dashed. In the second step all tiles drop to a second row: the three vowels slide together at the left inside a heavy outline, forming one block, and the consonants T, R, N, G, L, S follow, making 7 units. In the third step a bracket under the row reads 7 units: 7! orders, an arc over the block reads 3! inside, and the product 7! times 3! equals 30240 appears at the bottom.

Example 6.14NCERT Example 14 · p. 112

Vowels together, and not all together

TRIANGLES has 9 different letters. Count its arrangements in which (i) the vowels I, A, E are all together; (ii) the vowels are not all together.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.15NCERT Example 15 · p. 112

A toran of flowers

A toran for a doorway has 10 flowers strung in a line: 5 marigolds, 2 roses and 3 jasmines. Flowers of the same kind look identical. How many different patterns are possible?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.16NCERT Example 16 · p. 113

Four questions about ASSESSMENT

ASSESSMENT has 10 letters: S four times, E twice, and A, M, N, T once each. Find the number of arrangements of all its letters, and how many of them (i) begin with M; (ii) have all the vowels together; (iii) do not have all the vowels together; (iv) begin with A and end with T.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

How many arrangements does the word LEVEL have?

120603020

Show answer

Answer (C) 30

L and E each appear twice:

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

. Dividing by only one

2!

gives 60.

Practice · 13 questions

Practice 6.3: questions with answers

Short answerEasyNCERT Exercise 6.3, Q1
With the digits 1 to 7, each used at most once, how many 4-digit numbers are possible?
Show answer
Answer 840
$^{7} P_{4} = 7 \times 6 \times 5 \times 4 = 840.$
Short answerMediumNCERT Exercise 6.3, Q2
How many 3-digit numbers have no repeated digit?
Show answer
Answer 648
Hundreds: 9 choices (1 to 9). Tens: 9 (0 is now allowed, but the hundreds digit is used up). Units: 8. $9 \times 9 \times 8 = 648$ .
Short answerMediumNCERT Exercise 6.3, Q3
How many 3-digit odd numbers can be made from 2, 3, 4, 5, 6, 8, 9 if no digit is repeated?
Show answer
Answer 90
Units from 3, 5, 9: 3 ways; then 6 digits for the hundreds place and 5 for the tens: $3 \times 6 \times 5 = 90$ .
4
Short answerEasyNCERT Exercise 6.3, Q4
Using the digits 1 to 6 with no repetition, find the number of
1. 3-digit numbers that can be formed;
  Show answer
  Answer 120
  $^{6} P_{3} = 120$ .
2. those 3-digit numbers that are greater than 400.
  Show answer
  Answer 60
  Hundreds digit 4, 5 or 6: 3 ways, then $5 \times 4$ for the rest: 60.
Short answerEasyNCERT Exercise 6.3, Q5
A cricket squad of 11 must name a captain and a vice-captain, two different players. How many choices are there?
Show answer
Answer 110
$^{11} P_{2} = 11 \times 10 = 110$ .
Short answerMediumNCERT Exercise 6.3, Q6
Find $n$ if $^{n} P_{2} :^{n + 1} P_{3} = 1 : 12$ .
Show answer
Answer 11
$^{n + 1} P_{3} = (n + 1) n (n - 1)$ , so the ratio is $\frac{1}{n + 1} = \frac{1}{12}$ and $n = 11$ .
7
Short answerMediumNCERT Exercise 6.3, Q7
Find $r$ if
1. $^{4} P_{r} = 2^{4} P_{r - 1}$
  Show answer
  Answer 3
  $\frac{4!}{(4 - r)!} = \frac{2 \cdot 4!}{(5 - r)!}$ gives $5 - r = 2$ , so $r = 3$ .
2. $5^{4} P_{r} = 12^{5} P_{r - 1}$
  Show answer
  Answer 2
  Cancelling gives $(6 - r) (5 - r) = 12$ , so $r^{2} - 11 r + 18 = 0$ and $r = 2$ or $9$ . As $r \leq 4$ , $r = 2$ .
Short answerEasyNCERT Exercise 6.3, Q8
How many arrangements of all the letters of JUPITER are there, using each letter once?
Show answer
Answer 5040
Seven different letters: $7! = 5040$ .
9
Short answerMediumNCERT Exercise 6.3, Q9
BIRTHDAY has 8 different letters (treat Y as a consonant). Count the strings of letters from it, no letter used twice, that use
1. 3 letters at a time?
  Show answer
  Answer 336
  $^{8} P_{3} = 336$ .
2. all 8 letters?
  Show answer
  Answer 40320
  $8! = 40 320$ .
3. all 8 letters, with a vowel at each end?
  Show answer
  Answer 1440
  The vowels I and A take the two ends in $2!$ ways; the 6 consonants fill the middle in $6!$ ways: $2 \times 720 = 1440$ .
Short answerHardNCERT Exercise 6.3, Q10
In how many distinct arrangements of BOOKKEEPER are the three E's not all together?
Show answer
Answer 141120
Total: $\frac{10!}{2! 2! 3!} = 151 200$ (O twice, K twice, E three times). Glue EEE: 8 units with O and K twice, $\frac{8!}{2! 2!} = 10 080$ . So $151 200 - 10 080 = 141 120.$
11
Short answerHardNCERT Exercise 6.3, Q11
DEMOCRATS has 9 different letters. In how many of its arrangements
1. is D first and S last?
  Show answer
  Answer 5040
  The other 7 letters fill the middle: $7! = 5040$ .
2. are the vowels E, O, A all together?
  Show answer
  Answer 30240
  Block + 6 consonants = 7 units: $7! \times 3! = 30 240$ .
3. are there exactly 3 letters between D and S?
  Show answer
  Answer 50400
  D and S can sit in positions (1, 5), (2, 6), …, (5, 9): 5 pairs, in 2 orders. The other 7 letters fill the rest in $7!$ ways: $5 \times 2 \times 5040 = 50 400.$
Quick answerMedium
Assertion (A): The letters of NOON can be arranged in 6 ways. Reason (R): $n$ objects of which $p$ are alike of one kind and $q$ are alike of another kind can be arranged in $\frac{n!}{p! q!}$ ways.
Both A and R are true, and R explains ABoth A and R are true, but R does not explain AA is true, but R is falseA is false, but R is true
Show answer
Answer (A) Both A and R are true, and R explains A
$\frac{4!}{2! 2!} = 6$ , which is exactly R applied to two N's and two O's. Both are true and R explains A.
Quick answerHard
In how many arrangements of the letters of PERMIT does E come somewhere before I?
720240360120
Show answer
Answer (C) 360
Swapping E and I pairs the $6! = 720$ arrangements so that exactly one of each pair has E first. So the answer is $\frac{720}{2} = 360$ .

6.4 Combinations

Four friends, Ira, Jai, Kabir and Lata, must send two of them to an inter-school quiz. What counts is the pair that travels; nobody cares whose name was read out first. When only which objects are chosen matters, and not the order of choosing, we are counting combinations (संचय).

Definition

Combination

A choice of objects from a collection in which only which objects are chosen counts, not the order of choosing.

ExampleFrom Ira, Jai, Kabir and Lata, the pair Ira and Jai is one combination; Jai and Ira is the same one.

If the two had different jobs, say captain and deputy, order would matter and there would be $^{4} P_{2} = 12$ choices. As plain pairs there are only 6: IJ, IK, IL, JK, JL and KL. Each pair was counted twice among the 12, once in each order. We write $^{n} C_{r}$ for the count of all selections of $r$ objects out of $n$ different ones, so here $^{4} C_{2} = 6$ .

Combinations hide in many settings. In a league where each of 8 teams plays every other team once, a match is a pair of teams, so there are $^{8} C_{2}$ matches. Six points on a circle, no three in a line, give one chord for each pair of points and one triangle for each set of three points (Fig. 6.9).

2/3Each chord joins a pair of points: one chord for every 2-point selection.

Fig. 6.9A chord is a 2-point selection and a triangle is a 3-point selection: order does not matter in either.NCERT Fig. 6.3 · p. 115

A circle carries six labelled points, A to F, evenly spaced; text beside it says six points, no three in a line. In the second step all fifteen chords joining the points in pairs are drawn, and the side text says a chord is a pair of points, with 6C2 equals 15. In the third step the chords fade and the triangle ACE is shaded, with the text a triangle is a set of 3 points and 6C3 equals 20.

To find a formula, compare selections with arrangements. Fig. 6.10 lists every captain–vice-captain pair from five players P, Q, R, S, T: $^{5} P_{2} = 20$ ordered pairs. Each two-person team appears twice, as PQ and QP, so there are $\frac{20}{2!} = 10$ teams.

3/3Keep one triangle of the grid: $\frac{20}{2!} = 10$ teams.

Fig. 6.10Each team of two shows up twice among the ordered pairs, once in each order.

A grid with rows for the captain and columns for the vice-captain, both labelled P, Q, R, S, T. The diagonal cells are greyed out because one person cannot hold both posts, leaving 20 cells such as PQ and QP; the side text says 20 ordered pairs, 5P2 equals 20. In the second step the cells PQ and QP are highlighted and a dashed line marks the diagonal: they are the same team. In the third step the cells below the diagonal fade, leaving 10 teams, 5C2 equals 10.

Groups of three behave the same way: each 3-person group can be lined up in $3!$ orders, so arrangements outnumber groups by a factor of $3! = 6$ . In general:

Theorem 6.5Arrangements and selections

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

Dividing by $r!$ and using (6.3) gives the formula for combinations:

Combinations

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

(6.5)

InteractiveArrange or choose lab

A counting lab with three mode buttons: arrange without repeats, arrange with repeats allowed, and choose a group. Two sliders set the number of objects n, from 2 to 6 letters A to F, and the number of places or the group size r. A row of boxes shows the number of choices for each place, and a status panel gives the count with its formula: nPr as a falling product, n to the power r, or nCr as nPr divided by r factorial. Below, every outcome is listed as a chip, up to 120. In choose mode a checkbox opens each group into the r factorial arrangements it contains. Nothing moves until the learner acts.

Fig. 6.11Pick a mode and set

n

and

r

. In Choose a group, tick the box to open every selection into its

r!

orders.

All of them. With $r = n$ , $^{n} C_{n} = \frac{n!}{n! 0!} = 1$ : there is one way to take everything.
None of them. We agree that $^{n} C_{0} = 1$ : the empty selection is the only selection of no objects. The formula agrees, as $\frac{n!}{0! n!} = 1$ , so (6.5) holds for $0 \leq r \leq n$ .
Symmetry. Replacing $r$ by $n - r$ only swaps the two factorials in the denominator, so $^{n} C_{n - r} =^{n} C_{r}$ .
Equal values. If $^{n} C_{a} =^{n} C_{b}$ , then either $a = b$ or $a + b = n$ .

Symmetry

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

(6.6)

The symmetry has a plain meaning, shown in Fig. 6.12: every way of picking $r$ objects to keep also fixes the $n - r$ that stay out, and the other way round. Picking 2 laddoos from a box of 6 to eat now decides the 4 you keep for later, so $^{6} C_{2} =^{6} C_{4} = 15$ .

2/2… is the same as choosing the 4 to leave: $^{6} C_{2} =^{6} C_{4} = 15$ .

Fig. 6.12Every choice of what to take is also a choice of what to leave.

Two long trays are drawn one above the other. The upper tray, labelled take 2, holds laddoos 2 and 5; the lower tray, labelled leave 4, holds laddoos 1, 3, 4 and 6. In the first step the take tray has a bold outline. In the second step the leave tray has the bold outline instead and the two taken laddoos fade, showing that the four left behind form a selection too.

Theorem 6.6Pascal's rule

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

Because $r! = r (r - 1)!$ and $(n - r + 1)! = (n - r + 1) (n - r)!$ both hold, the two terms share one factor. Call it $K = \frac{n!}{(r - 1)! (n - r)!}$ for short.

Then $^{n} C_{r} = \frac{K}{r}$ and $^{n} C_{r - 1} = \frac{K}{n - r + 1}$ , so their sum is $K$ times a sum of two small fractions. Over a common denominator they add up neatly:

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

Multiply by $K$ . Using $r (r - 1)! = r!$ and $(n - r + 1) (n - r)! = (n - r + 1)!$ once more, the sum becomes

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

and the right side is $^{n + 1} C_{r}$ by (6.5).

A counting argument gives the same rule without algebra. From $n + 1$ people, one of them Meera, choose $r$ . Either Meera is in the group, and the other $r - 1$ come from the remaining $n$ people in $^{n} C_{r - 1}$ ways, or she is not, and all $r$ come from the other $n$ in $^{n} C_{r}$ ways. The cases do not overlap, so they add.

Write the numbers $^{n} C_{r}$ in rows, one row for each $n$ , and Pascal's rule says that each inner entry is the sum of the two entries above it (Fig. 6.13). Indian scholars knew this triangle long before Pascal: the commentator Halayudha described it as the Meru prastara in the 10th century.

3/3Every inner entry of row 5 is built the same way.

Fig. 6.13Each inner entry is the sum of the two entries just above it.

Rows of numbered boxes form a triangle, labelled n equals 0 to n equals 5 on the left: 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; and a faint last row 1 5 10 10 5 1. In the second step the boxes 4 and 6 in row 4 are highlighted and two arrows point down to the box 10 in row 5, with the label 4C1 plus 4C2 equals 5C2. In the third step the whole of row 5 becomes solid.

Example 6.17NCERT Example 17 · p. 117

Using the symmetry

If $^{n} C_{10} =^{n} C_{6}$ , find $n$ and then $^{n} C_{14}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 6.18NCERT Example 18 · p. 117

A school panel

A 4-member panel is to be picked from 4 teachers and 5 students. How many different panels are possible? How many of them have exactly 2 teachers and 2 students?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.19NCERT Example 19 · p. 118

Three cards from a pack

Three cards are drawn together from a standard pack of 52 (see the box above). Find the number of possible hands, and how many of them (i) are all of one suit; (ii) have three different suits; (iii) are all face cards; (iv) have two red cards and one black card; (v) are all of one colour.

Try it yourself first. Then reveal the solution one step at a time.

Flash cards · 6

Arrange or choose?

1 / 6

Practice · 11 questions

Practice 6.4: questions with answers

Short answerEasyNCERT Exercise 6.4, Q1
If $^{n} C_{7} =^{n} C_{4}$ , find $^{n} C_{3}$ .
Show answer
Answer 165
$n = 7 + 4 = 11$ , so $^{11} C_{3} = \frac{11 \cdot 10 \cdot 9}{6} = 165.$
2
Short answerMediumNCERT Exercise 6.4, Q2
Find $n$ if
1. $^{2 n} C_{2} :^{n} C_{2} = 22 : 5$
  Show answer
  Answer 6
  The ratio is $\frac{2 (2 n - 1)}{n - 1}$ . Setting it equal to $\frac{22}{5}$ gives $20 n - 10 = 22 n - 22,$ so $n = 6$ .
2. $^{2 n} C_{2} :^{n} C_{2} = 5 : 1$
  Show answer
  Answer 3
  $\frac{2 (2 n - 1)}{n - 1} = 5$ gives $4 n - 2 = 5 n - 5$ , so $n = 3$ .
3
Short answerEasyNCERT Exercise 6.4, Q3
Fifteen points lie on a circle.
1. How many chords join them in pairs?
  Show answer
  Answer 105
  $^{15} C_{2} = 105$ .
2. How many triangles have all three corners among these points?
  Show answer
  Answer 455
  $^{15} C_{3} = 455$ .
Short answerEasyNCERT Exercise 6.4, Q4
A team of 2 boys and 4 girls is to be picked from 6 boys and 7 girls. How many such teams are possible?
Show answer
Answer 525
$^{6} C_{2} \times^{7} C_{4} = 15 \times 35 = 525.$
Short answerMediumNCERT Exercise 6.4, Q5
A basket holds 5 different mangoes, 4 different guavas and 6 different bananas. In how many ways can you pick 6 fruits, 2 of each kind?
Show answer
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
How many 5-card hands from a pack of 52 contain exactly two aces?
Show answer
Answer 103776
Choose 2 of the 4 aces and 3 of the 48 other cards: $^{4} C_{2} \times^{48} C_{3} = 6 \times 17 296 = 103 776.$
Short answerMediumNCERT Exercise 6.4, Q7
A hockey squad has 16 players, 3 of whom are goalkeepers. A team of 11 must include exactly one goalkeeper. How many teams are possible?
Show answer
Answer 858
$^{3} C_{1} \times^{13} C_{10} = 3 \times 286 = 858.$
Short answerEasyNCERT Exercise 6.4, Q8
A jar has 7 green and 4 yellow marbles, all distinguishable. In how many ways can 3 green and 2 yellow marbles be taken out?
Show answer
Answer 210
$^{7} C_{3} \times^{4} C_{2} = 35 \times 6 = 210.$
Short answerEasyNCERT Exercise 6.4, Q9
A student must choose 6 electives out of 12, and 2 particular electives are compulsory. In how many ways can the choice be made?
Show answer
Answer 210
Choose the other 4 from the remaining 10: $^{10} C_{4} = 210$ .
Quick answerMedium
$^{10} C_{3} +^{10} C_{4}$ equals
$^{20} C_{7}$ $^{10} C_{7} = 120$ $^{11} C_{4} = 330$ $^{11} C_{3} = 165$
Show answer
Answer (C) $^{11} C_{4} = 330$
$^{10} C_{4} +^{10} C_{3} =^{11} C_{4} = 330.$ Check: $120 + 210 = 330$ .
Quick answerMedium
Assertion (A): From 10 books, choosing 3 to take on a trip can be done in as many ways as choosing 7 to leave at home. Reason (R): $^{n} C_{r} =^{n} C_{n - r}$ .
Both A and R are true, and R explains ABoth A and R are true, but R does not explain AA is true, but R is falseA is false, but R is true
Show answer
Answer (A) Both A and R are true, and R explains A
Each choice of 3 to take fixes the 7 left behind, so $^{10} C_{3} =^{10} C_{7} = 120$ . R is the general form of A.

Miscellaneous Examples

The remaining examples mix the tools of this chapter: select and then arrange, split into cases, count the complement, move through a dictionary list, and keep objects apart with gaps.

Example 6.20NCERT Example 20 · p. 119

Select first, then arrange

From the letters of EDUCATION, each used at most once, how many 4-letter words (meaningful or not) can you make that contain exactly 2 vowels and 2 consonants?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.21NCERT Example 21 · p. 120

At least: by cases and by the complement

A committee of 4 is chosen from 5 women and 6 men. In how many ways can it be formed if it has (i) no woman; (ii) at least one woman and at least one man; (iii) at least 3 women?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.22NCERT Example 22 · p. 121

The 50th word in a dictionary list

All arrangements of the letters of PAPER are listed in dictionary order. How many words are in the list, and what is the 50th word?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.23NCERT Example 23 · p. 121

Large numbers from repeated digits

How many numbers greater than 500 000 can be formed by using all of the digits 5, 0, 3, 5, 7, 0, each exactly as often as it appears in the list?

Try it yourself first. Then reveal the solution one step at a time.

To keep certain objects apart, seat the others first and use the gaps between them (Fig. 6.14).

3/3Put the 3 girls in 3 different gaps, in order: $^{5} P_{3} = 60$ ways. No two girls are neighbours.

Fig. 6.14Girls placed in different gaps can never sit next to each other.NCERT Example 24 · p. 122

Four square tiles marked B stand in a row with five dashed gap boxes around them, one at each end and three in between; text says boys first: 4! orders. In the second step the gaps are outlined boldly and marked with crosses, three round tiles marked G wait above, and the text says 5 gaps for the girls. In the third step the three girls drop into the first, third and fourth gaps, and the text says girls in 3 of the 5 gaps.

Example 6.24NCERT Example 24 · p. 122

No two girls side by side

In how many ways can 4 boys and 3 girls sit in a row of 7 chairs so that no two girls sit next to each other?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

How many 3-letter words with 1 vowel and 2 consonants can be made from the letters of TIGER?

Show answer

Answer 36

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

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

letter sets, each arranged in

3! = 6

ways: 36.

Practice · 14 questions

Practice questions with answers (NCERT “Miscellaneous Exercise on Chapter 6”)

Short answerMediumNCERT Miscellaneous Exercise, Q1
Using the letters of FRACTION, each at most once, how many 4-letter words (meaningful or not) contain exactly 1 vowel and 3 consonants?
Show answer
Answer 720
Vowels A, I, O; consonants 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
How many arrangements of all the letters of COUNTRIES keep the vowels together and the consonants together?
Show answer
Answer 5760
Vowels O, U, I, E form one block ( $4!$ orders inside) and consonants C, N, T, R, S another ( $5!$ orders). The blocks come in $2!$ orders: $2 \times 24 \times 120 = 5760.$
3
Short answerMediumNCERT Miscellaneous Exercise, Q3
A committee of 6 is to be formed from 7 men and 5 women. Count the possible committees that have
1. exactly 2 women?
  Show answer
  Answer 350
  $^{5} C_{2} \times^{7} C_{4} = 10 \times 35 = 350.$
2. at least 3 women?
  Show answer
  Answer 462
  3, 4 or 5 women: $10 \cdot 35 + 5 \cdot 21 + 1 \cdot 7 = 462.$
3. at most 1 woman?
  Show answer
  Answer 112
  0 or 1 woman: $^{7} C_{6} +^{5} C_{1} \times^{7} C_{5} = 7 + 105 = 112.$
Short answerHardNCERT Miscellaneous Exercise, Q4
All arrangements of the letters of BALLOON are listed in dictionary order. How many words come before the first word that starts with L?
Show answer
Answer 360
Starting with A, the letters B, L, L, O, O, N give $\frac{6!}{2! 2!} = 180$ words. Starting with B, the letters A, L, L, O, O, N give another 180. So 360 words come first.
Short answerHardNCERT Miscellaneous Exercise, Q5
How many 5-digit numbers divisible by 5 can be made from the digits 0, 2, 3, 5, 6, 8 if no digit is repeated?
Show answer
Answer 216
Last digit 0: arrange 4 of the other 5 digits in front, $^{5} P_{4} = 120$ . Last digit 5: the first digit is 2, 3, 6 or 8 (4 ways), then $^{4} P_{3} = 24$ for the middle, giving 96. Total $120 + 96 = 216$ .
Short answerMediumNCERT Miscellaneous Exercise, Q6
Of the 26 letters of the English alphabet, 5 are vowels. Count the 4-letter strings that use one vowel and three different consonants.
Show answer
Answer 159600
$^{5} C_{1} \times^{21} C_{3} \times 4! = 5 \times 1330 \times 24 = 159 600.$
Short answerMediumNCERT Miscellaneous Exercise, Q7
A test has Part A and Part B with 6 questions each. A candidate must answer 7 questions, with at least 2 from each part. How many different sets of 7 questions can she pick?
Show answer
Answer 780
Splits (A, B) = (2, 5), (3, 4), (4, 3), (5, 2): $15 \cdot 6 + 20 \cdot 15 + 15 \cdot 20 + 6 \cdot 15 = 780.$
Short answerMediumNCERT Miscellaneous Exercise, Q8
How many 4-card hands from a pack of 52 contain exactly one heart?
Show answer
Answer 118807
$^{13} C_{1} \times^{39} C_{3} = 13 \times 9139 = 118 807.$
Short answerHardNCERT Miscellaneous Exercise, Q9
Four boys and 3 girls sit in a row of 7 chairs numbered 1 to 7. The girls may sit only on odd-numbered chairs. How many seatings are possible?
Show answer
Answer 576
The odd chairs are 1, 3, 5, 7. The girls take 3 of these 4 chairs in order: $^{4} P_{3} = 24$ . The boys fill the 4 chairs left: $4! = 24$ . Total $24 \times 24 = 576$ .
Short answerHardNCERT Miscellaneous Exercise, Q10
From a class of 20, 8 students are to go on a field trip. Two of them are twins who will go only together. How many groups are possible?
Show answer
Answer 62322
Twins go: choose 6 more from the other 18, $^{18} C_{6} = 18 564$ . Twins stay: choose all 8 from the 18, $^{18} C_{8} = 43 758$ . Total $62 322$ .
Short answerMediumNCERT Miscellaneous Exercise, Q11
In how many arrangements of the letters of POSSESSIONS are all the S's together?
Show answer
Answer 2520
POSSESSIONS has 11 letters: S five times, O twice, and P, E, I, N. The S block with P, O, O, E, I, N makes 7 units with O twice: $\frac{7!}{2!} = 2520$ .
Quick answerMedium
The number of diagonals of a decagon (10-sided polygon) is
45904035
Show answer
Answer (D) 35
$^{10} C_{2} = 45$ segments join pairs of vertices; 10 of them are sides, leaving 35 diagonals.
Quick answerHard
How many rectangles of all sizes can be found on an $8 \times 8$ chessboard?
Show answer
Answer 1296
A rectangle is fixed by choosing 2 of the 9 horizontal lines and 2 of the 9 vertical lines: $^{9} C_{2} \times^{9} C_{2} = 36 \times 36 = 1296.$
Quick answerHard
Assertion (A): $^{10} C_{4} =^{10} C_{6}$ . Reason (R): $^{n} P_{r} =^{n} P_{n - r}$ for every $r$ .
Both A and R are true, and R explains ABoth A and R are true, but R does not explain AA is true, but R is falseA is false, but R is true
Show answer
Answer (C) A is true, but R is false
A is true by symmetry (both equal 210). R is false: $^{10} P_{4} = 5040$ but $^{10} P_{6} = 151 200$ . Order makes the two counts differ.

Recap

Permutations and Combinations: key points and summary

Fundamental principle of counting: tasks done one after another in $m_{1}, m_{2}, \dots, m_{k}$ ways can be done together in $m_{1} m_{2} \dots m_{k}$ ways (Section 6.2).
Successive stages multiply, disjoint cases add, and the most restricted place is filled first (Example 6.3).
Permutations count orderings; $^{n} P_{r} = n (n - 1) \dots (n - r + 1)$ (Theorem 6.1).
$n! = 1 \times 2 \times \dots \times n,$ $n! = n (n - 1)!$ and $0! = 1$ (Section 6.3.2).
$^{n} P_{r} = \frac{n!}{(n - r)!}$ for $0 \leq r \leq n$ , so $^{n} P_{n} = n!$ and $^{n} P_{0} = 1$ ((6.3)).
With repetition allowed, filling $r$ places from $n$ objects gives $n^{r}$ arrangements (Theorem 6.2).
With $p_{1}, p_{2}, \dots, p_{k}$ alike objects of each kind, $n$ objects have $\frac{n!}{p_{1}! p_{2}! \dots p_{k}!}$ arrangements (Theorem 6.4).
Together: glue a block and multiply by the orders inside it; not all together: total minus together (Example 6.14).
A combination is a selection in which order does not matter, and $^{n} P_{r} =^{n} C_{r} \times r!$ (Theorem 6.5).
$^{n} C_{r} = \frac{n!}{r! (n - r)!}$ , with $^{n} C_{0} =^{n} C_{n} = 1$ and $^{n} C_{n - r} =^{n} C_{r}$ ((6.5)).
Pascal's rule: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ (Theorem 6.6).
Mixed problems: select then arrange, split at least into cases or use the complement, and keep objects apart with gaps (Miscellaneous Examples).

Timeline

Historical note: counting across cultures

Ancient India
Sushruta's 63 tastes
The medical text Sushruta Samhita, traditionally dated to about the 6th century BCE, says that 6 basic tastes give 63 mixtures taken one, two or more at a time. Check: $^{6} C_{1} +^{6} C_{2} + \dots +^{6} C_{6} = 63.$
c. 3rd century BCE
Pingala counts metres
Pingala's Chhandas Shastra gave rules for listing and counting the patterns of long and short syllables in a line of verse.
Early Jain texts
Vikalpa
Jain scholars treated the counting of selections and arrangements, which they called vikalpa, as a subject in its own right.
Ancient China
Hexagrams of the I Ching
The I Ching (Book of Changes) uses figures of six lines, each whole or broken: $2^{6} = 64$ hexagrams in all.
c. 850
Mahavira's rule
Mahavira's Ganita Sara Sangraha stated a general rule for the number of combinations, equivalent to the formula for $^{n} C_{r}$ .
10th century
Meru prastara
Halayudha, commenting on Pingala, described the triangle of numbers now named after Pascal.
c. 1140
Abraham ibn Ezra
He counted the possible conjunctions of the seven planets then known, taken two, three or more at a time: 120 in all.
1150
Bhaskara II's Lilavati
A chapter called Anka Pasha (net of numbers) works out arrangements of digits, including digits that repeat.
1321
Levi ben Gerson
His Maaseh Hoshev proved formulas for permutations and combinations with an early form of mathematical induction.
1713
Bernoulli's Ars Conjectandi
Jacob Bernoulli's book, published after his death, gave the first complete treatment of permutations and combinations.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
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	Miscellaneous Exercise on Chapter 6 · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous Exercise on Chapter 6 · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous Exercise on Chapter 6 · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous Exercise on Chapter 6 · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous Exercise on Chapter 6 · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous Exercise on Chapter 6 · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous Exercise on Chapter 6 · Q7
NCERT Miscellaneous Exercise, Q8	Miscellaneous Exercise on Chapter 6 · Q8
NCERT Miscellaneous Exercise, Q9	Miscellaneous Exercise on Chapter 6 · Q9
NCERT Miscellaneous Exercise, Q10	Miscellaneous Exercise on Chapter 6 · Q10
NCERT Miscellaneous Exercise, Q11	Miscellaneous Exercise on Chapter 6 · Q11

Quick check: MCQs on Permutations and Combinations

6.2 Fundamental Principle of Counting4 questions
6.3.2 Factorial notation4 questions
6.3.3 Derivation of the formula for nPr2 questions
6.3.4 Permutations when all the objects are not distinct4 questions
6.4 Combinations4 questions
Miscellaneous Examples3 questions

Glossary · 7

Key terms (English–Hindi)

addition principle: If the outcomes split into cases that cannot happen together, the total is the sum of the case counts.First seen in the chapter →

combinationसंचयsanchay: A selection of objects in which order does not matter; $^{n} C_{r}$ counts those of $r$ out of $n$ different objects.First seen in the chapter →

factorialक्रमगुणितkramgunit: $n! = 1 \times 2 \times \dots \times n$ for a natural number $n$ , with $0! = 1$ .
fundamental principle of countingगणना का आधारभूत सिद्धांतganana ka aadharbhoot siddhant: If one task can be done in $m$ ways and then a second in $n$ ways, the two in order can be done in $m n$ ways.First seen in the chapter →

Pascal's rule: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ : each inner entry of Pascal's triangle is the sum of the two above it.
permutationक्रमचयkramchay: An arrangement of some or all of a set of objects in a definite order; $^{n} P_{r}$ counts those of $r$ out of $n$ different objects.First seen in the chapter →

tree diagramवृक्ष आरेखvriksh aarekh: A branching picture of successive choices in which each path from the start to a leaf is one outcome.First seen in the chapter →

6.1 Introduction

6.2 Fundamental Principle of Counting

6.3 Permutations

6.3.1 Permutations when all the objects are distinct

6.3.2 Factorial notation

6.3.3 Derivation of the formula for nPr

6.3.4 Permutations when all the objects are not distinct

6.4 Combinations

Miscellaneous Examples

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