If P (2n − 1, n) : P (2n + 1, n − 1) = 22 : 7 find n.
Question:

If P (2n − 1, n) : P (2n + 1, n − 1) = 22 : 7 find n.

Solution:

P (2n − 1, n):P (2n + 1, n − 1) = 22:7

$\Rightarrow \frac{(2 n-1) !}{(2 n-1-n) !} \times \frac{(2 n+1-n+1) !}{(2 n+1) !}=\frac{22}{7}$

 

$\Rightarrow \frac{(2 n-1) !}{(n-1) !} \times \frac{(n+2) !}{(2 n+1) !}=\frac{22}{7}$

$\Rightarrow \frac{(2 n-1) !}{(n-1) !} \times \frac{(n+2)(n+1)(n)(n-1) !}{(2 n+1)(2 n)(2 n-1) !}=\frac{22}{7}$

$\Rightarrow \frac{(n+2)(n+1)(n)}{(2 n+1)(2 n)}=\frac{22}{7}$

 

$\Rightarrow \frac{(n+2)(n+1)}{2(2 n+1)}=\frac{22}{7}$

$\Rightarrow 7 n^{2}+21 n+14=88 n+44$

$\Rightarrow 7 n^{2}-67 n-30=0$

$\Rightarrow 7 n^{2}-70 n+3 n-30=0$

 

$\Rightarrow(n-10)(7 n+3)=0$

$\therefore n=10$ or $\frac{-3}{7}$

Since $n$ cannot be negative, it is equal to10.

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