PSEB 11th Class Maths Solutions Chapter 7 Permutations and Combinations Ex 7.2

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Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 7 Permutations and Combinations Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.2

Question 1.
Evaluate
(i) 8!
(ii) 4! – 3!
Answer.
(i) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320

(ii) 4! = 4 × 3 × 2 × 1 = 24
3! = 3 × 2 × 1=6
∴ 4! – 3! = 24 – 6 = 18

Question 2.
Is 3! + 4! = 7!?
Answer.
3! = 3 × 2 × 1 = 6
4! = 4 × 3 × 2 × 1 =2 4
∴ 3! + 4! = 6 + 24 = 30
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040
3! + 4! ≠ 7!.

PSEB 11th Class Maths Solutions Chapter 7 Permutations and Combinations Ex 7.2

Question 3.
Compute \(\frac{8 !}{6 ! \times 2 !}\)
Answer.
\(\frac{8 !}{6 ! \times 2 !}=\frac{8 \times 7 \times 6 !}{6 ! \times 2 \times 1}=\frac{8 \times 7}{2}\) = 28.

Question 4.
If \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\), find x.
Answer.

PSEB 11th Class Maths Solutions Chapter 7 Permutations and Combinations Ex 7.2 1

PSEB 11th Class Maths Solutions Chapter 7 Permutations and Combinations Ex 7.2

Question 5.
Evaluate \(\) when
(i) n = 6, r = 2
(ii) n = 9, r = 5
Answer.
(i) When n = 6, r = 2 then

= \(\frac{n !}{(n-r) !}=\frac{6 !}{(6-2) !}=\frac{6 !}{4 !}\)

= \(\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}\)
= 6 × 5 = 30

(ii) When n = 9, r = 5, then
= \(\frac{n !}{(n-r) !}=\frac{9 !}{(9-5) !}=\frac{9 !}{4 !}\)

= \(\frac{9 \times 5 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}\)
= 9 × 8 × 7 × 6 × 5 = 15120.

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