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Question
If P (2n − 1, n) : P (2n + 1, n − 1) = 22 : 7 find n.
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Solution
P (2n − 1, n):P (2n + 1, n − 1) = 22:7
\[\Rightarrow \frac{\left( 2n - 1 \right)!}{\left( 2n - 1 - n \right)!} \times \frac{\left( 2n + 1 - n + 1 \right)!}{\left( 2n + 1 \right)!} = \frac{22}{7}\]
\[ \Rightarrow \frac{\left( 2n - 1 \right)!}{\left( n - 1 \right)!} \times \frac{\left( n + 2 \right)!}{\left( 2n + 1 \right)!} = \frac{22}{7}\]
\[ \Rightarrow \frac{\left( 2n - 1 \right)!}{\left( n - 1 \right)!} \times \frac{\left( n + 2 \right)\left( n + 1 \right)\left( n \right)\left( n - 1 \right)!}{\left( 2n + 1 \right)\left( 2n \right)\left( 2n - 1 \right)!} = \frac{22}{7}\]
\[ \Rightarrow \frac{\left( n + 2 \right)\left( n + 1 \right)\left( n \right)}{\left( 2n + 1 \right)\left( 2n \right)} = \frac{22}{7}\]
\[ \Rightarrow \frac{\left( n + 2 \right)\left( n + 1 \right)}{2\left( 2n + 1 \right)} = \frac{22}{7}\]
\[ \Rightarrow 7 n^2 + 21n + 14 = 88n + 44\]
\[ \Rightarrow 7 n^2 - 67n - 30 = 0\]
\[ \Rightarrow 7 n^2 - 70n + 3n - 30 = 0\]
\[ \Rightarrow \left( n - 10 \right)\left( 7n + 3 \right) = 0\]
\[ \therefore n = 10 or \frac{- 3}{7}\]
\[\text{Sincencannot be negative, it is equal to10}.\]
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