Question 3 & 4 Exercise 6.1

Solutions of Question 3 & 4 of Exercise 6.1 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Prove that $\dfrac{1}{6 !}+\dfrac{2}{7 !}+\dfrac{3}{8 !}=\dfrac{75}{8 !}$

We are taking the L.H.S of the above given equation. $$\begin{align}\dfrac{1}{6 !}+\dfrac{2}{7 !}+\dfrac{3}{8 !}&=\dfrac{1}{6 !}+\dfrac{2}{7.6 !}+\dfrac{3}{8.7 .6 !} \\ & =\dfrac{56+16+3}{8 !}\\ &=\dfrac{75}{8 !}\end{align}$$

Prove that $\dfrac{(n+5) !}{(n+3) !}=n^2+9 n+20$

We are taking the L.H.S of the above given equation. $$\begin{align}\dfrac{(n+5) !}{(n+3) !}&=\dfrac{(n+5)(n+4)(n+3) !}{(n+3) !} \\ & =(n+5)(n+4)\\ &=n^2+9 n+20\end{align}$$

Find the value of $n$, when $\dfrac{n(n !)}{(n-5) !}=\dfrac{12(n !)}{(n-4) !}$

We are given: $$\begin{align}\dfrac{n(n !)}{(n-5) !}&=\dfrac{12(n !)}{(n-4) !} \\ \Rightarrow \dfrac{n}{(n-5) !}&=\dfrac{12}{(n-4)(n-5) !} \\ \Rightarrow n&=\dfrac{12}{(n-4)} \\ \Rightarrow n(n-4)&=12 \\ \rightarrow n^2-4 n-12&=0 \\ \rightarrow n^2+2 n-6 n \quad 12&=0 \\ \Rightarrow n(n+2)-6(n+2)&=0 \\ \Rightarrow(n-6)(n+2)&=0 \\ \Rightarrow \text { Either. } n&=-2 \text { or } n=6\end{align}$$ $n$ can not b negative, therefore $n=6$.

Find the value of $n$, when $\dfrac{n !}{(n-4) !}: \dfrac{(n-1) !}{(n-4) !}=9: 1$

We are given: $$\begin{align} \dfrac{n !}{(n-4) !}: \dfrac{(n-1) !}{(n-4) !}&=9: 1 \\ \Rightarrow \dfrac{n !}{(n-4) !} \times \dfrac{(n-4) !}{(n-1) !}&=9 \\ \Rightarrow \dfrac{n !}{(n-1) !}&=9 \\ \Rightarrow \dfrac{n(n-1) !}{(n-1) !}&=9 \\ \Rightarrow n&=9 \end{align}$$

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