Permutation and Combination Class 11 NCERT Solutions Chapter 7 | Complete Solutions

Permutation and Combination Class 11 NCERT Solutions provides clear and step-by-step answers to all the questions of Chapter 7. This chapter helps students understand important concepts like factorials, permutations, combinations, and their applications in solving counting problems. On this page, you will find detailed explanations of every exercise to help you prepare for school exams and competitive exams with confidence.

Permutation and Combination Class 11 NCERT Solutions

EXERCISE 7.1

1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5
   assuming that
   (i) repetition of the digits is allowed?
   (ii) repetition of the digits is not allowed?

Sol:

(i) When repetition of digits is allowed

Each place (Hundreds, Tens, Units) can be filled with any of the $5$ digits.$$\text{Hundreds place} = 5 \text{ choices}$$$$\text{Tens place} = 5 \text{ choices}$$$$\text{Units place} = 5 \text{ choices}$$$$\text{Total numbers} = 5 \times 5 \times 5 = 5^3 = 125$$$$\boxed{125}$$

(ii) When repetition of digits is NOT allowed

$$\text{Hundreds place} = 5 \text{ choices}$$$$\text{Tens place} = 4 \text{ choices}$$$$\text{Units place} = 3 \text{ choices}$$$$\text{Total numbers} = 5 \times 4 \times 3 = 60$$

This can also be written as:$${}^{5}P_{3} = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 60$$

2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the
digits can be repeated?

Sol:

Step 1: Condition for Even Number

A number is even if its units digit is even.

Even digits available:$$2, 4, 6$$

So,$$\text{Units place} = 3 \text{ choices}$$

Step 2: Hundreds Place

Since repetition is allowed, any of the 6 digits can be used.$$\text{Hundreds place} = 6 \text{ choices}$$

Step 3: Tens Place

Again, repetition is allowed.$$\text{Tens place} = 6 \text{ choices}$$

Step 4: Total Number of 3-digit Even Numbers

$$\text{Total numbers} = 6 \times 6 \times 3$$$$= 108$$

3. How many 4-letter code can be formed using the first 10 letters of the English
alphabet, if no letter can be repeated?

Sol: First $10$ letters of the English alphabet:$$A, B, C, D, E, F, G, H, I, J$$

We need to form 4-letter codes with no repetition.

Since order matters in a code, we use permutations.

Step 1: Number of choices for each position

$$\text{1st position} = 10 \text{ choices}$$$$\text{2nd position} = 9 \text{ choices}$$$$\text{3rd position} = 8 \text{ choices}$$$$\text{4th position} = 7 \text{ choices}$$

Step 2: Total number of codes

$$10 \times 9 \times 8 \times 7$$$$= 5040$$

Using permutation formula

$${}^{10}P_{4} = \frac{10!}{(10-4)!}$$$$= \frac{10!}{6!}$$$$= 10 \times 9 \times 8 \times 7$$$$= 5040$$

4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if
each number starts with 67 and no digit appears more than once?

Sol:

Digits available:$$0,1,2,3,4,5,6,7,8,9$$

The telephone number:

• Has 5 digits
• Starts with 67
• No digit is repeated

Step 1: Fix the first two digits

Since the number must start with $67$67,$$\text{First two digits are fixed.}$$First two digits are fixed.

Digits already used: $6, 7$

Remaining digits available:$$\{0,1,2,3,4,5,8,9\}$$

Total remaining digits = $8$

Step 2: Fill the remaining 3 positions

Since repetition is not allowed:$$\text{3rd position} = 8 \text{ choices}$$$$\text{4th position} = 7 \text{ choices}$$$$\text{5th position} = 6 \text{ choices}$$

Step 3: Total number of telephone numbers

$$8 \times 7 \times 6$$$$= 336$$

Using permutation formula

$${}^{8}P_{3} = \frac{8!}{(8-3)!}$$$$= \frac{8!}{5!} = 8 \times 7 \times 6$$$$= 336$$

5. A coin is tossed 3 times and the outcomes are recorded. How many possible
outcomes are there?

Sol:

When a coin is tossed, there are 2 possible outcomes:$$H \quad \text{(Head)} \qquad T \quad \text{(Tail)}$$

The coin is tossed 3 times.

Step 1: Number of choices for each toss

$$\text{1st toss} = 2 \text{ outcomes}$$$$\text{2nd toss} = 2 \text{ outcomes}$$$$\text{3rd toss} = 2 \text{ outcomes}$$

Step 2: Total number of possible outcomes

$$2 \times 2 \times 2 = 2^3$$$$= 8$$

Sample Space

$$\{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \}$$

6. Given 5 flags of different colours, how many different signals can be generated if
each signal requires the use of 2 flags, one below the other?

Sol:

We are given:

• $5$ flags of different colours
• Each signal uses $2$2 flags
• One flag is placed below the other

Since the order (top and bottom) matters, we use permutations.

Step 1: Number of choices

$$\text{Top position} = 5 \text{ choices}$$$$\text{Bottom position} = 4 \text{ choices}$$

(The bottom flag must be different from the top flag.)

Step 2: Total number of signals

$$5 \times 4 = 20$$

Using permutation formula

$${}^{5}P_{2} = \frac{5!}{(5-2)!}$$$$= \frac{5!}{3!} = 5 \times 4$$$$= 20$$

EXERCISE 7.2

Evaluate
(i) 8 ! (ii) 4 ! – 3 !

Is 3 ! + 4 ! = 7 ! ?

Compute 8!/6! 2!

If 6!1​+7!1​=8!x​, find $x$
We know:
$$7! = 7 \times 6!, \quad 8! = 8 \times 7 \times 6!$$$$\frac{1}{6!} = \frac{7}{7!}$$
$$\frac{1}{6!} + \frac{1}{7!} = \frac{7}{7!} + \frac{1}{7!} = \frac{8}{7!}$$$$\frac{8}{7!} = \frac{8 \times 8}{8!} = \frac{64}{8!}$$
$$\frac{x}{8!}$$$$\boxed{x = 64}$$5. Evaluate
$$\frac{n!}{(n-r)!}$$(i) When $n = 6, r = 2$n=6,r=2
$$\frac{6!}{(6-2)!} = \frac{6!}{4!}$$$$= \frac{6 \times 5 \times 4!}{4!} = 6 \times 5 = \boxed{30}$$(ii) When $n = 9, r = 5$n=9,r=5
$$\frac{9!}{(9-5)!} = \frac{9!}{4!}$$$$= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!}$$$$= 9 \times 8 \times 7 \times 6 \times 5$$

EXERCISE 7.3

1. Question : How many 3-digit numbers can be formed using the digits 000 to 999 if no digit is repeated?

Solution

For a 3-digit number, the first digit cannot be $0$.$$\text{Hundreds place} = 9 \text{ choices } (1\text{–}9)$$

After choosing the first digit:$$\text{Tens place} = 9 \text{ choices}$$$$\text{Units place} = 8 \text{ choices}$$$$\text{Total numbers} = 9 \times 9 \times 8$$$$= 648$$$$\boxed{648}$$

2. Question : How many 4-digit numbers are there in which no digit is repeated?

Solution

Digits available: $0$to $9$$$\text{Thousands place} = 9 \text{ choices } (1\text{–}9)$$$$\text{Hundreds place} = 9 \text{ choices}$$$$\text{Tens place} = 8 \text{ choices}$$$$\text{Units place} = 7 \text{ choices}$$$$\text{Total numbers} = 9 \times 9 \times 8 \times 7$$$$= 4536$$$$\boxed{4536}$$

3. Question :How many 3-digit even numbers can be formed using the digits 1,2,3,4,6,71,2,3,4,6,71,2,3,4,6,7, if no digit is repeated?

Solution

Even digits available:$$2,4,6$$$$\text{Units place} = 3 \text{ choices}$$

After fixing the units digit:$$\text{Hundreds place} = 5 \text{ choices}$$$$\text{Tens place} = 4 \text{ choices}$$$$\text{Total numbers} = 3 \times 5 \times 4$$$$= 60$$$$\boxed{60}$$

4. Question Find the number of 4-digit numbers that can be formed using the digits 1,2,3,4,51,2,3,4,51,2,3,4,5, if no digit is repeated. How many of these numbers will be even?

Solution

(i) Total 4-digit numbers

$${}^{5}P_{4} = 5 \times 4 \times 3 \times 2$$$$= 120$$$$\boxed{120}$$

(ii) Even numbers among them

Even digits available:$$2,4$$$$\text{Units place} = 2 \text{ choices}$$

Remaining digits $= 4$$$\text{Thousands place} = 4 \text{ choices}$$$$\text{Hundreds place} = 3 \text{ choices}$$$$\text{Tens place} = 2 \text{ choices}$$$$\text{Total even numbers} = 2 \times 4 \times 3 \times 2$$$$= 48$$

5. Question : From a committee of 8 persons, in how many ways can a chairman and a vice-chairman be chosen, assuming that one person cannot hold more than one position?

Solution

Since the posts are different, order matters.$$\text{Chairman} = 8 \text{ choices}$$$$\text{Vice-chairman} = 7 \text{ choices}$$$$\text{Total ways} = 8 \times 7 = 56$$$$\boxed{56}$$

8. Question : How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

The word EQUATION has 8 distinct letters.$$\text{Total arrangements} = 8!$$Total arrangements=8! $$= 40320$$$$\boxed{40320}$$

9. Question : How many words, with or without meaning, can be formed from the letters of the word MONDAY, assuming that no letter is repeated, if:

The word MONDAY has 6 distinct letters.

(i) 4 letters are used at a time

$${}^{6}P_{4} = \frac{6!}{(6-4)!}$$$$= \frac{6!}{2!} = 6 \times 5 \times 4 \times 3$$$$= 360$$$$\boxed{360}$$

(ii) All letters are used at a time

$$6! = 720$$$$\boxed{720}$$

(iii) All letters are used but the first letter is a vowel

Vowels in MONDAY:$$O, A$$$$\text{First position} = 2 \text{ choices}$$

Remaining letters can be arranged in:$$5! \text{ ways}$$$$= 2 \times 5! = 2 \times 120 = 240$$$$\boxed{240}$$

10. Question : In how many distinct permutations of the letters of the word MISSISSIPPI do the four I’s not come together?

Letters in MISSISSIPPI:$$M=1,\; I=4,\; S=4,\; P=2$$

Total permutations:

$$\frac{11!}{4!4!2!}$$$$= 34650$$

Case when four I’s come together

Treat the four I’s as one unit.

Total objects:$$1(M) + 1(I\text{-block}) + 4(S) + 2(P) = 8$$$$\text{Arrangements} = \frac{8!}{4!2!}$$$$= 840$$

Required number

$$34650 – 840 = 33810$$$$\boxed{33810}$$

11. Question : In how many ways can the letters of the word PERMUTATIONS be arranged if:

Letters in PERMUTATIONS:

Total letters = 12

Repetitions:$$T=2,\quad O=2$$

Total arrangements:$$\frac{12!}{2!2!}$$

(i) The word starts with P and ends with S

Fix P at first and S at last.

Remaining letters = 10

Repetitions remain $T=2,\; O=2$$$\text{Arrangements} = \frac{10!}{2!2!}$$$$= 907200$$$$\boxed{907200}$$

(ii) All the vowels are together

Vowels: $E, U, A, I, O, O$

Treat vowels as one block.

Consonants: $P, R, M, T, T, N, S$

Now total units:$$7 \text{ consonants} + 1 \text{ vowel block} = 8$$

Arrange these:$$\frac{8!}{2!}$$

(2 T’s repeated)

Arrange vowels inside block:$$\frac{6!}{2!}$$

(2 O’s repeated)

Total:$$\frac{8!}{2!} \times \frac{6!}{2!}$$$$= 10080 \times 360$$$$= 3628800$$$$\boxed{3628800}$$

(iii) There are always 4 letters between P and S

Total positions = 12

If there are 4 letters between P and S, their positions differ by 5.

Possible position pairs:$$(1,6), (2,7), (3,8), (4,9), (5,10), (6,11), (7,12)$$

Total = 7 ways

P and S can interchange:$$7 \times 2 = 14$$

Arrange remaining 10 letters:$$\frac{10!}{2!2!}$$$$= 907200$$

Total:$$14 \times 907200$$$$= 12700800$$

EXERCISE 7.4

1. If$${^nC_8} = {^nC_2},$$

find the value of$${^nC_2}.$$

Sol:

We know the property of combinations:$${}^{n}C_{r} = {}^{n}C_{n-r}$$

Given,$${}^{n}C_{8} = {}^{n}C_{2}$$

This is possible only when:$$8 = n – 2$$$$n = 10$$

Verification

$${}^{10}C_{8} = {}^{10}C_{2}$$$${}^{10}C_{2} = \frac{10 \times 9}{2} = 45$$$${}^{10}C_{8} = {}^{10}C_{2} = 45$$

2. Determine the value of $n$n if:

(i)$${^{2n}C_3} : {^nC_3} = 12 : 1$$

(ii)$${^{2n}C_3} : {^nC_3} = 11 : 1$$

Solution

We know:$${}^{n}C_{3} = \frac{n(n-1)(n-2)}{6}$$$${}^{2n}C_{3} = \frac{2n(2n-1)(2n-2)}{6}$$

(i)

$$\frac{{}^{2n}C_{3}}{{}^{n}C_{3}} = 12$$$$\frac{\frac{2n(2n-1)(2n-2)}{6}}{\frac{n(n-1)(n-2)}{6}} = 12$$

Cancel $6$$$\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 12$$

Since $2n-2 = 2(n-1)$$$= \frac{2n(2n-1)\cdot 2(n-1)}{n(n-1)(n-2)}$$

Cancel $n$ and $(n-1)$$$= \frac{4(2n-1)}{n-2}$$

So,$$\frac{4(2n-1)}{n-2} = 12$$$$4(2n-1) = 12(n-2)$$$$8n – 4 = 12n – 24$$$$20 = 4n$$$$n = 5$$$$\boxed{n=5}$$

(ii)

$$\frac{{}^{2n}C_{3}}{{}^{n}C_{3}} = 11$$

Using the simplified expression:$$\frac{4(2n-1)}{n-2} = 11$$$$4(2n-1) = 11(n-2)$$$$8n – 4 = 11n – 22$$$$18 = 3n$$$$n = 6$$$$\boxed{n=6}$$

3. How many chords can be drawn through 21 points on a circle?

Sol:

To find how many chords can be drawn through 21 points on a circle:

A chord is formed by joining any two points on the circle.

So, the number of chords = Number of ways to choose 2 points from 21 points.$$\text{Number of chords} = ^{21}C_2$$$$^{21}C_2 = \frac{21 \times 20}{2 \times 1}$$$$= \frac{420}{2}$$$$= 210$$

4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Sol:

To form a team of 3 boys and 3 girls from 5 boys and 4 girls:

We select separately:

• 3 boys from 5 boys
• 3 girls from 4 girls

Step 1: Select boys

$$^{5}C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$$

Step 2: Select girls

$$^{4}C_{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$$

Step 3: Total ways

$$\text{Total ways} = 10 \times 4 = 40$$

5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls, and 5 blue balls, if each selection consists of 3 balls of each colour.

Sol:

We are given:

• 6 red balls
• 5 white balls
• 5 blue balls
• Total balls to be selected = 9
• Condition: 3 balls of each colour

Step 1: Select red balls

Number of ways to select 3 red balls from 6:$$^{6}C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$

Step 2: Select white balls

Number of ways to select 3 white balls from 5:$$^{5}C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$$

Step 3: Select blue balls

Number of ways to select 3 blue balls from 5:$$^{5}C_{3} = 10$$

Step 4: Total number of ways

$$\text{Total ways} = 20 \times 10 \times 10$$$$= 2000$$

6. Determine the number of 5-card combinations from a deck of 52 cards if there is exactly one ace in each combination.

Sol :

We want the number of 5-card combinations from a standard 52-card deck with exactly one ace.

A deck has:

• 4 Aces
• 48 non-ace cards

Step 1: Choose 1 Ace from 4

$$^{4}C_{1} = 4$$

Step 2: Choose remaining 4 cards from 48 non-aces

$$^{48}C_{4}$$$$^{48}C_{4} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1}$$$$= 194580$$

Step 3: Total combinations

$$\text{Total ways} = 4 \times 194580$$$$= 778320$$

7. In how many ways can a cricket team of 11 players be selected from 17 players, in which only 5 players can bowl, if each team must include exactly 4 bowlers?

Sol: We are given:

• Total players = 17
• Bowlers = 5
• Therefore, non-bowlers = 17 − 5 = 12
• Team size = 11
• Exactly 4 bowlers must be included

Step 1: Select 4 bowlers from 5

$$^{5}C_{4} = 5$$

Step 2: Select remaining 7 players from 12 non-bowlers

$$^{12}C_{7}$$$$^{12}C_{7} = ^{12}C_{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}$$$$= 792$$

Step 3: Total number of teams

$$\text{Total ways} = 5 \times 792$$$$= 3960$$

8. A bag contains 5 black balls and 6 red balls. Determine the number of ways in which 2 black balls and 3 red balls can be selected.

Sol:

We are given:

• 5 black balls
• 6 red balls
• Selection required: 2 black balls and 3 red balls

Step 1: Select 2 black balls from 5

$$^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10$$

Step 2: Select 3 red balls from 6

$$^{6}C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$

Step 3: Total number of ways

$$\text{Total ways} = 10 \times 20$$$$= 200$$

9. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Sol:

We are given:

• Total available courses = 9
• A student must choose 5 courses
• 2 specific courses are compulsory

Step 1: Fix the compulsory courses

Since 2 courses are compulsory, they are already included in the selection.

So, we now need to choose:$$5 – 2 = 3 \text{ more courses}$$

Step 2: Remaining courses

After removing the 2 compulsory courses:$$9 – 2 = 7 \text{ courses left}$$

Now select 3 courses from these 7:$$^{7}C_{3}$$$$^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1}$$$$= 35$$

For the official Class 11 Physics Solutions, you can visit:

1. NCERT Textbooks

ncert resources

Class-wise Solutions

Class 12:

Class 12 Physics – NCERT Solutions

Class 12 Chemistry – NCERT Solutions

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• Class 11 Physics – NCERT Solutions
• Class 11 Chemistry – NCERT Solutions
• Class 11 Biology – NCERT Solutions
• Class 11 Math – NCERT Solutions

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• Class 10 Science – NCERT Solutions
• Class 10 Math – NCERT Solutions
• Class 10 Science – CBSE SQP 2024