Question 1 Exercise 6.3

Solutions of Question 1 of Exercise 6.3 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.

Solve $^n C_2=36$ for $n$.

We are given: $$\begin{align}&^n C_2=36\\ & \Rightarrow \dfrac{n !}{(n-2) ! 2 !}=36 \\ & \Rightarrow \dfrac{n(n-1)(n-2) !}{(n-2) ! \cdot 2}=36 \\ & \Rightarrow n(n-1)=72 \\ & \Rightarrow n^2-n-72=0 \\ & \Rightarrow n^2-9 n+8 n-72=0\\ & \Rightarrow n(n-9)+8(n-9)=0 \\ & \Rightarrow(n-9)(n+8)=0 \\ & \Rightarrow \text { either } n=9 \text { or } n=-8\end{align}$$ But $n$ can not be negative therefore, we have $n=9$.

Solve $^{n+1} C_4=6,^{n-1} C_2$ for $n$.

We are given: $$\begin{align} & { }^{n+1} C_4=6 .^{n-1} C_2 \\ & \Rightarrow \dfrac{(n+1) !}{(n+1-4) ! 4 !}=6\dfrac{(n-1) !}{(n-1-2) ! 2 !} \\ & \Rightarrow \dfrac{(n+1) n(n-1) !}{(n-3) ! 4 !}=6 \cdot \dfrac{(n-1) !}{(n-3) ! 2 !} \\ & \Rightarrow \dfrac{n(n+1)}{4 !}=\dfrac{6}{2 !} \\ & \Rightarrow \dfrac{n(n+1)}{12}=6 \\ & \Rightarrow n^2+n-72=0 \\ & \Rightarrow n^2+9 n-8 n-72=0 \\ & \Rightarrow n(n+9)-8(n+9)=0 \\ & \Rightarrow(n+9)(n-8)=0 \\ & \Rightarrow \text { either } n=-9 \text { or } n=8 \end{align}$$ But $n$ can not be negative, therefore we have $n=8$.

Solve $n^2 C_2=30 .{ }^n C_3$ for $n$.

We are given: $$\begin{align} & { }^2 C_2=30 .^n C_3 \\ & \Rightarrow \dfrac{(n^2) !}{(n^2-2) ! 2 !}=30 \dfrac{n !}{(n-3) ! 3 !} \\ & \Rightarrow \dfrac{n^2(n^2-1)(n^2-2) !}{(n^2-2) ! 2 !}=30 \cdot \dfrac{n(n-1)(n-2)(n-3) !}{(n-3) ! 3 \cdot 2 !} \\ & \Rightarrow n^2(n^2-1)=10 n(n-1)(n-2) \\ & \Rightarrow n^2(n-1)(n+1)=10 n(n-1)(n-2) \\ & \Rightarrow n(n+1)=10(n-2) \\ & \Rightarrow n^2+n-10 n+20=0\\ & \Rightarrow n^2-9 n+20=0 \\ & \Rightarrow n^2-4 n-5 n+20=0 \\ & \Rightarrow n(n-4)-5(n-4)=0 \\ & \Rightarrow(n-4)(n-5)=0 \\ & \Rightarrow \text { Either } n=4, \text { or } n=5 \end{align}$$

Question 2 >