Exercise 7.2 ncert solutions maths class 11

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Exercise 7.2(Permutations and Combinations)

Exercise 7.2 ncert solutions maths class 11

Question 1: Evaluate

(i) 8!

(ii) 4! – 3!

Solution: (i) Given  8!

We know that 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 40320

(ii) Given 4!-3!

4! – 3! = (4 × 3!) – 3!

= 3! (4-1)

= 3 × 2 × 1 × 3

= 18

Question 2: Is 3! + 4! = 7!?

Solution:  LHS = 3! + 4!

3! + 4! = (3 × 2 × 1) + (4 × 3 × 2 × 1)

= 6 + 24

= 30

Taking R.H.S.

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Therefore LHS ≠ RHS

Therefore 3! + 4! ≠ 7!

3. Compute \dfrac{8!}{6!\times 2!}

Solution: Given \dfrac{8!}{6!\times 2!}

\Rightarrow \dfrac{8\times 7 \times 6!}{6!\times 2\times 1}

\Rightarrow \dfrac{8\times 7}{2}

\Rightarrow 28

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

Solution: Given, 

\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}

\Rightarrow \frac{1}{6!}+\frac{1}{7\times 6!}=\frac{x}{8\times 7!}

\Rightarrow \frac{1}{6!}(1+\frac{1}{7})= \frac{x}{8\times 7\times 6!}

\Rightarrow \frac{1}{6!}\times \frac{8}{7}= \frac{x}{8\times 7\times 6!}

\Rightarrow 8 = \frac{x}{8}

\Rightarrow x = 64

5. Evaluate \dfrac{n!}{(n-r)!},

When
(i) n = 6, r = 2
(ii) n = 9, r = 5

Solution: (i) Given n = 6 and r = 2

Putting the value of n and r we get

\dfrac{6!}{(6-2)!}

\Rightarrow \dfrac{6!}{4!}

\Rightarrow \dfrac{6\times 5\times 4!}{4!}

\Rightarrow 6\times 5

\Rightarrow 30

(i) Given n = 9 and r = 5

Putting the value of n and r we get

\dfrac{9!}{(9-5)!}

\Rightarrow \dfrac{9!}{4!}

\Rightarrow \dfrac{9\times 8\times 7\times 6 \times 5 \times 4!}{4!}

\Rightarrow 9\times 8\times 7\times 6 \times 5

\Rightarrow 15120

 

Chapter 7 Permutation and Combination Ncert math solution

Exercise 7.1 Permutation and Combination Ncert math solution

Exercise 7.2 Permutation and Combination Ncert math solution

Exercise 7.3 Permutation and Combination Ncert math solution

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