Question 2 Review Exercise 6

Solutions of Question 2 of Review Exercise 6 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.

If ${ }^{2 n} C_r={ }^{2 n} C_{r+2}$; find $r$.

Given that: $$\begin{align} { }^{2 n} C_r&={ }^{2 n} C_{r+2} \\ \Rightarrow \dfrac{(2 n) !}{(2 n-r) ! r !}&=\dfrac{(2 n) !}{(2 n-(r+2)) !(r+2) !}\end{align}$$ Dividing both sides by $(2 n)$ ! we get $$\begin{align} \Rightarrow \dfrac{1}{(2 n-r) ! r !}&=\dfrac{1}{(2 n-r-2) !(r+2) !} \\ \Rightarrow \dfrac{1}{(2 n-r)(2 n-r-1)(2 n-r-2) ! r !}&=\dfrac{1}{(2 n-r-2) !(r+2) !} \\ \Rightarrow \dfrac{1}{(2 n-r)(2 n-r-1) r !}& =\dfrac{1}{(r+2)(r+1) r !} \\ \Rightarrow(r+2)(r+1)&=(2 n-r)(2 n-r-1) \\ \Rightarrow r^2+3 r+2&= 4 n^2-2 n r-2 n-2 n r+r^2+r \\ \Rightarrow 3 r+2&=4 n^2-4 n r-2 n+r \\ \Rightarrow 4 n r+3 r-r&=4 n^2-2 n \\ \Rightarrow 4 n r+2 r&=2(2 n^2-n) \\ \Rightarrow 2 r(2 n+1)&=2(2 n^2-n)\\ \Rightarrow r&=\dfrac{2 n^2-n}{2 n+1}\end{align}$$

If ${ }^{18} C_r={ }^{18} C_{r+2}$; find ${ }^r C_5$.

Given that: $$\begin{align} { }^{18} C_r&={ }^{18} C_{r+2} \\ \Rightarrow \dfrac{(18) !}{(18-r) ! r !}&= \dfrac{18 !}{[18-(r+2)] !(r+2) !}\end{align}$$ Dividing both sides by $18!$ $$\begin{align}\dfrac{1}{(18-r) ! r !}&= \dfrac{1}{(16-r) !(r+2)(r+1) r !} \\ \Rightarrow \dfrac{1}{(18-r)(17-r)(16-r) !}& =\dfrac{1}{(16-r) !(r+2)(r+1)} \\ \Rightarrow(r+2)(r+1)&=(18-r)(17-r) \\ \Rightarrow r^2+3 r+2&=306-18 r-17 r+r^2 \\ \Rightarrow 3 r+2&=306-35 r \\ \Rightarrow 38 r&=304 \\ \Rightarrow r&=\dfrac{304}{38}=8 . \\ \text { Now }{ }^r C_5&={ }^8 C_5=\dfrac{8 !}{(8-5) ! 5 !}=56\end{align}$$

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