Exercise 6.1 (Solutions)

The solutions of the Exercise 6.1 of book “Model Textbook of Mathematics for Class XI” published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan are given on this page. This exercise consists of the question related to factorial function.

The book misses the definition of the factorial function, which is defined as follows:

Question 1. Evaluate the following:
(i) $10!$ (ii) $\dfrac{12!}{7! 3! 2!}$ (iii) $\dfrac{4!-2!}{3!+5!}$ (iv) $\dfrac{(n-1)!}{(n-2)!}$ (v) $\dfrac{8!}{(6!)^2}$
Solution: Question 1

Question 2. Write the following in factorial form:
(i) 14.13 .12 .11 (ii) 1.3.5.7.9 (iii) $n\left(n^{2}-1\right)$ (iv) $\frac{(n-3)(n-2)(n-1)}{n(n-4)}$
Solution: Question 2

Question 3. Prove the following:
(i) $\frac{1}{5!}+\frac{3}{6!}+\frac{1}{7!}=\frac{4}{315}$ (ii) $\frac{(n-1)!}{(n-3)!}=n^{2}-3 n+2$
Solution: Question 3 & 4

Question 4. Show that:
(i) $\frac{(2 n)!}{n!}=2^{n}(1.3 .5 \ldots(2 n-1))$ (ii) $\frac{(2 n-1)!}{n!}=2^{n-1}(1.3 .5 \ldots(2 n-1))$
Solution: Question 3 & 4

Question 5. Find the values of $n$ in the following.
(i) $\frac{n}{(n-4)!}=\frac{3.3!}{(n-3)!}$ (ii) $\frac{n!}{(n-4)!}: \frac{(n-1)!}{(n-3)!}=36: 2$
Solution: Question 5

Question 6(i-v). Prove the following for $n \in \mathbb{N}$,
(i) $(2 n)!=2^{n}(n!)[1.3 .5 \ldots(2 n-1)]$ (ii) $(n+1)[n!n+(n-1)!(2 n-1)+(n-2)!(n-1)!]=(n+2)$ !
(iii) $\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}=\frac{(n+1)!}{r!(n-r+1)!}$ (iv) $\frac{n!}{r!}=n(n-1)(n-2) \ldots(r+1)$ (v) $\quad(n-r+1) \cdot \frac{n!}{(n-r+1)!}=\frac{n!}{(n-r)!}$
Solution: Question 6(1-v)

Question 6(vi-ix). Prove the following for $n \in \mathbb{N}$,
(vi) 33 ! is divisible by $2^{15}$ (vii) $\frac{2 n!}{[(n-1)!]^{2}}=\frac{n(n+1)(n+2) \ldots(2 n-1)(2 n)}{(n-1)!}$
(viii) $(n!+1)$ is not divisible by any natural number between 2 and $n$. (ix) $\quad(n!)^{2} \leq n^{n} . n!<(2 n)$ !
Solution: Question 6(vi-ix)

Question 7(i-vi). Find $n$, if
(i) $\frac{n!}{(n-2)!}=930, n \geq 2$ (ii) $\frac{n!}{(n-5)!}=20 \cdot \frac{n!}{(n-3)!}, n \geq 5$ (iii) $(n+2)!=60 \cdot(n-1)$ !
(iv) $(n+2)!=132 . n$ ! (v) $\quad(n+2)!=56 . n!$ (vi) $\frac{1}{9!}+\frac{1}{10!}=\frac{n}{11!}$
Solution: Question 7(i-vi)

Question 7(vii-xi). \item Find $n$, if
(vii) $n!=990 .(n-3)$ ! (viii) $(n+1)!=6 .(n-1)$ ! (ix) $\frac{(n+2)!}{(2 n-1)!}=\frac{(n+3)!}{(2 n+1)!} \frac{72}{7}$
(x) $\frac{(2 n)!}{4!(2 n-3)!} \cdot \frac{4!(n-4)!}{n!}=52$ (xi) $\frac{1}{2(n-2)!}: \frac{1}{4!(n-4)!}=2, n \geq 4$
Solution: Question 7(vii-xi)