NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.3

NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.3 Permutations and Combinations help students understand the concept of combinations, where the order of selection does not matter. This exercise focuses on selecting objects rather than arranging them.

Prepared according to the latest CBSE Class 11 Maths syllabus, Exercise 6.3 covers problems based on selecting items using the combination formula. Regular practice of this exercise helps students improve logical thinking and accuracy in solving selection-based questions.

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.3

Permutations and Combinations

Medium

Q.

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?

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Permutations and Combinations

Easy

Q.

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

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Permutations and Combinations

Easy

Q.

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

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Permutations and Combinations

Medium

Q.

How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

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Permutations and Combinations

Medium

Q.

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

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Permutations and Combinations

Easy

Q.

From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?

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Permutations and Combinations

Easy

Q.

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

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Permutations and Combinations

Difficult

Q.

In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

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Permutations and Combinations

Medium

Q.

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

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Permutations and Combinations

Medium

Q.

In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?

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The solutions are explained in a clear and step-by-step format so students can easily apply the correct formulas and solve problems confidently in exams.

1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Ans: We must form a 3-digit numbers using digits 1–9. That means, the order of the digits matters.

Hence, there will be as many 3-digit numbers as there are permutations of 9 digits taken 3 at a time.

So, required number of 3-digit numbers
= ⁹P₃

$$= \frac{9!}{(9-3)!} = \frac{9!}{6!}$$

=(9−3)!9!​=6!9!​

$$= \frac{9 \times 8 \times 7 \times 6!}{6!}$$

=6!9×8×7×6!​

$$= 9 \times 8 \times 7 = 504$$

=9×8×7=504

Hence, there are 504 3-digit numbers formed.

2. How many 4-digit numbers are there with no digit repeated?

Ans: The thousands place of the 4-digit number is to be filled with any of the digits from 1 to 9 as the digit 0 cannot be included. Therefore, the number of ways in which thousands place can be filled is 9. The hundreds, tens, and units place can be filled by any of the digits from 0 to 9. However, the digits cannot be repeated in the 4-digit numbers and thousands place is already occupied with a digit. The hundreds, tens, and units place is to be filled by the remaining 9-digits. Therefore, there will be as many such 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Number of such 3-digit numbers
= ⁹P₃

$$= \frac{9!}{(9-3)!} = \frac{9!}{6!}$$

=(9−3)!9!​=6!9!​

$$= \frac{9 \times 8 \times 7 \times 6!}{6!}$$

=6!9×8×7×6!​

$$= 9 \times 8 \times 7 = 504$$

=9×8×7=504

Thus, by multiplication principle, the required number of 4-digit numbers are

$$9 \times 504 = 4536$$

9×504=4536

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

Ans: We have to form 3-digit even numbers using the given six digits, 1, 2, 3, 4, 6, and 7, without repeating the digits. Then, units can be filled in 3 ways by any of the digits, 2, 4 or 6. Since the digits cannot be repeated in the 3-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining 5-digits. Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5-digits is the permutation of 5 different digits taken 2 at a time.

So, the number of ways of filling hundreds and tens place is,

= ⁵P₂

$$= \frac{5!}{(5-2)!} = \frac{5!}{3!}$$

=(5−2)!5!​=3!5!​

$$= \frac{5 \times 4 \times 3!}{3!}$$

=3!5×4×3!​

$$= 5 \times 4 = 20$$

=5×4=20

Thus, by multiplication principle, the required number of 3-digit numbers are

$$3 \times 20 = 60$$

3×20=60

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

Ans: We must form 4-digit numbers using the digits, 1, 2, 3, 4, and 5.

So, there will be as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time.

That is, required number of 4-digit numbers
= ⁵P₄

$$= \frac{5!}{(5-4)!} = \frac{5!}{1!}$$

=(5−4)!5!​=1!5!​

$$= 1 \times 2 \times 3 \times 4 \times 5$$

=1×2×3×4×5

$$= 120$$

=120

Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5 even numbers end with either 2 or 4. The number of ways in which units place is filled with digits is 2. Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining 4-digits.

Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at time.

Number of ways of filling the remaining places
= ⁴P₃

$$= \frac{4!}{(4-3)!}$$

=(4−3)!4!​

$$= 4!$$

=4!

$$= 4 \times 3 \times 2 \times 1 = 24$$

=4×3×2×1=24

Thus, by multiplication principle, the required number of even numbers is

$$24 \times 2 = 48$$

24×2=48

5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

Ans: We must choose a chairman and a vice chairman from a committee of 8 persons in such a way that one person cannot hold more than one position. Here, the number of ways of choosing a chairman and a vice chairman is the permutation of 8 different objects taken 2 at a time.

Thus, required number of ways
= ⁸P₂

$$= \frac{8!}{(8-2)!}$$

=(8−2)!8!​

$$= \frac{8 \times 7 \times 6!}{6!}$$

=6!8×7×6!​

$$= 8 \times 7$$

=8×7

$$= 56$$

=56

6. Find n if ⁿ⁻¹P₃ : ⁿP₄ = 1 : 9.

Ans: Here,

ⁿ⁻¹P₃ : ⁿP₄ = 1 : 9

$$\frac{^{n-1}P_3}{^{n}P_4} = \frac{1}{9}$$

nP4​n−1P3​​=91​

$$\frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9}$$

(n−4)!n!​(n−1−3)!(n−1)!​​=91​

$$\frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = \frac{1}{9}$$

(n−4)!(n−1)!​×n!(n−4)!​=91​

$$\frac{(n-1)!}{n \times (n-1)!} = \frac{1}{9}$$

n×(n−1)!(n−1)!​=91​

$$\frac{1}{n} = \frac{1}{9}$$

n1​=91​

Hence, n = 9.

7. Find r if:

(i) ⁵Pᵣ = 2 × ⁶Pᵣ₋₁

Ans:

$$\frac{5!}{(5-r)!} = 2 \times \frac{6!}{(6-r+1)!}$$

(5−r)!5!​=2×(6−r+1)!6!​

$$\frac{5!}{(5-r)!} = \frac{2 \times 6 \times 5!}{(7-r)(6-r)(5-r)!}$$

(5−r)!5!​=(7−r)(6−r)(5−r)!2×6×5!​

$$(7-r)(6-r) = 12$$

(7−r)(6−r)=12

$$42 - 6r - 7r + r^2 = 12$$

42−6r−7r+r2=12

$$r^2 - 13r + 30 = 0$$

r2−13r+30=0

$$(r-3)(r-10) = 0$$

(r−3)(r−10)=0

$$r = 3 \text{ or } 10$$

r=3 or 10

It is known that

$$^nP_r = \frac{n!}{(n-r)!}, \text{ where } 0 \le r \le n$$

nPr​=(n−r)!n!​, where 0≤r≤n

$$0 \le r \le 5$$

0≤r≤5

Hence, r = 3.

(ii) ⁵Pᵣ = ⁶Pᵣ₋₁

Ans:

$$\frac{5!}{(5-r)!} = \frac{6!}{(6-r+1)!}$$

(5−r)!5!​=(6−r+1)!6!​

$$\frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)(6-r)(5-r)!}$$

(5−r)!5!​=(7−r)(6−r)(5−r)!6×5!​

$$(7-r)(6-r) = 6$$

(7−r)(6−r)=6

$$42 - 6r - 7r + r^2 = 6$$

42−6r−7r+r2=6

$$r^2 - 13r + 36 = 0$$

r2−13r+36=0

$$(r-4)(r-9) = 0$$

(r−4)(r−9)=0

It is known that

$$^nP_r = \frac{n!}{(n-r)!}, \text{ where } 0 \le r \le n$$

nPr​=(n−r)!n!​, where 0≤r≤n

$$0 \le r \le 5$$

0≤r≤5

Hence, r = 4.

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

Ans: There are 8 different letters in the word EQUATION. Therefore, the number of words that can be formed using all the letters of the word EQUATION, using each letter exactly once, is the number of permutations of 8 different objects taken 8 at a time, which is ⁸P₈ = 8!.

Thus, required number of words that can be formed is

$$8! = 40320$$

8!=40320

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

(i) 4 letters are used at a time,

Ans: There are 6 different letters in the word MONDAY.

Number of 4-letter words that can be formed from the letters of the word MONDAY, without repetition of letters, is the number of permutations of 6 different objects taken 4 at a time, which is ⁶P₄.

Thus, required number of words that can be formed using 4 letters at a time are

$$^6P_4 = \frac{6!}{(6-4)!}$$

6P4​=(6−4)!6!​

$$= \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!}$$

=2!6×5×4×3×2!​

$$= 6 \times 5 \times 4 \times 3$$

=6×5×4×3

$$= 360$$

=360

(ii) all letters are used at a time,

Ans: Number of words that can be formed by using all the letters of the word MONDAY at a time is the number of permutations of 6 different objects taken 6 at a time, which is ⁶P₆ = 6!.

Thus, required number of words that can be formed when all letters are used at a time

$$6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6$$

6!=1×2×3×4×5×6

$$= 720$$

=720

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

Ans: In the given word, there are 2 different vowels, which must occupy the rightmost place of the words formed. This can be done only in 2 ways. Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining 5 letters. This can be done in 5! ways.

Thus, in this case, required number of words that can be formed is

$$5! \times 2 = 120 \times 2$$

5!×2=120×2

$$= 240$$

=240

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

Ans: In the given word MISSISSIPPI, I appear 4 times, S appears 4 times, P appears 2 times, and M appears just once. Therefore, number of distinct permutations of the letters in the given word

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

4!4!2!11!​

$$= \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4 \times 3 \times 2 \times 1 \times 2 \times 1}$$

=4×3×2×1×2×111×10×9×8×7×6×5×4!​

$$= \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1 \times 2 \times 1}$$

=4×3×2×1×2×111×10×9×8×7×6×5​

$$= 34650$$

=34650

There are 4 I’s in the given word. When they occur together, they are treated as single object [IIII] for the time being. This single object together with the remaining 7 objects will account for 8 objects.

These 8 objects in which there are 4 S’s, and 2 P’s can be arranged in

$$\frac{8!}{4!2!}$$

4!2!8!​

ways that means, 840 ways.

Number of arrangements where all I’s occurred together is in 840 ways.

Thus, number of distinct permutations of the letters in MISSISSIPPI in which four I’s do not come together is

$$34650 - 840 = 33810$$

34650−840=33810

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

(i) Words start with P and end with S,

Ans:
In the word PERMUTATIONS, there are 2 T’s, and all the other letters appear only once.

If P and S are fixed at the extreme ends (P at the left end and S at the right end), then 10 letters are left.

Hence, in this case, required number of arrangements

$$\frac{10!}{2!} = 1814400$$

2!10!​=1814400

(ii) Vowels are all together,

Ans: There are 5 vowels in the given word, each appearing only once.

Since they must always occur together, they are treated as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects, in total.

These 8 objects in which there are 2 T’s can be arranged in

$$\frac{8!}{2!}$$

2!8!​

ways.

Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5! ways. Therefore, by multiplication principle, required number of arrangements in this case

$$\frac{8!}{2!} \times 5! = 2419200$$

2!8!​×5!=2419200

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

Ans: The letters must be arranged in such a way that there are always 4 letters between P and S. Therefore, in a way, the places of P and S are fixed. The remaining 10 letters in which there are 2 T’s can be arranged in

$$\frac{10!}{2!}$$

2!10!​

ways.

Since, possible places of P & S are 1 & 6, 2 & 7, 3 & 8, 4 & 9, 5 & 10, 6 & 11, 7 & 12. Then there can be 4 letters between P & S. Also, P & S can be interchanged as it won’t affect the number of letters between them.

So, the letters P and S can be placed such that there are 4 letters between them in

$$2 \times 7 = 14$$

2×7=14

ways. Therefore, by multiplication principle, required number of arrangements in this case

$$\frac{10!}{2!} \times 14 = 25401600$$

2!10!​×14=25401600

FAQs – Class 11 Maths Chapter 6 Exercise 6.3 Permutations and Combinations

Q1. What is the focus of Exercise 6.3?
Exercise 6.3 focuses on combinations, which involve selecting objects where the order does not matter.

Q2. What is the formula for combinations?
The number of combinations of n distinct objects taken r at a time is:
nCr = n! / [r! (n − r)!]

Q3. What is the difference between permutation and combination?

• Permutation: Order matters

• Combination: Order does not matter

Q4. Why is Exercise 6.3 important for exams?
Combination-based questions are frequently asked in board and competitive exams and are considered scoring when concepts are clear.

Q5. How can students prepare effectively for Exercise 6.3?
Students should practice selection-based problems, understand when to use combination formulas, and revise factorial concepts thoroughly.