Question 5, Exercise 6.1

Solutions of Question 5 of Exercise 6.1 of Unit 06: Permutation and Combination. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the value of $n$: $\quad \dfrac{n}{(n-4)!}=\dfrac{3.3!}{(n-3)!}$

Solution. $$\begin{align*} \dfrac{n}{(n-4)!}&=\dfrac{3.3!}{(n-3)!}\\ \dfrac{n}{(n-4)!}&=\dfrac{3.3!}{(n-3)(n-4)!}\\ n&=\dfrac{3\times 6}{n-3}\\ n(n-3)&=18\\ n^2-3n&=18\\ n^2-2n-18&=0\\ n^2+3n-6n-18&=0\\ n(n+3)-6(n+3)&=0\\ (n+3)(n-6)&=0\\ n+3=0\quad \text{or}&\quad n-6=0\\ n=-3\quad\text{or}&\quad n=6 \end{align*}$$ The value of $n$ is not negative, so $n=6$

Find the value of $n$: $\quad \dfrac{n!}{(n-4)!}:\dfrac{(n-1)!}{(n-3)!}=36:2$

Solution. $$\begin{align*} \dfrac{n!}{(n-4)!}:\dfrac{(n-1)!}{(n-3)!}&=36:2\\ \dfrac{n(n-1)!}{(n-4)!}:\dfrac{(n-1)!}{(n-3)(n-4)!}&=36:2\\ \dfrac{n}{1}:\dfrac{1}{(n-3)}&=36:2\\ n(n-3)&=\dfrac{36}{2}\\ n^2-3n-18&=0\\ n^2+3n-6n-18&=0\\ n(n+3)-6(n+3)&=0\\ (n+3)(n-6)&=0\\ n+3=0\quad \text{or}&\quad n-6=0\\ n=-3\quad \text{or}&\quad n=6\\ \end{align*}$$ $n$ cannot be negative, so $n=6$

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