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Question
If \[\frac{(2n)!}{3! (2n - 3)!}\] and \[\frac{n!}{2! (n - 2)!}\] are in the ratio 44 : 3, find n.
Solution
\[\frac{(2n)!}{3!(2n - 3)!}: \frac{n!}{2!(n - 2)!} = 44: 3 \]
\[ \Rightarrow \frac{(2n)!}{3!(2n - 3)!} \times \frac{2!(n - 2)!}{n!} = \frac{44}{3}\]
\[ \Rightarrow \frac{(2n)(2n - 1)(2n - 2) [(2n - 3)!]}{3(2!)(2n - 3)!} \times \frac{2!(n - 2)!}{n(n - 1) [(n - 2)!]} = \frac{44}{3}\]
\[ \Rightarrow \frac{(2n)(2n - 1)(2n - 2)}{3} \times \frac{1}{n(n - 1)} = \frac{44}{3}\]
\[ \Rightarrow \frac{(2n)(2n - 1)(2)(n - 1)}{3} \times \frac{1}{n(n - 1)} = \frac{44}{3}\]
\[ \Rightarrow \frac{4(2n - 1)n(n - 1)}{3} \times \frac{1}{n(n - 1)} = \frac{44}{3}\]
\[ \Rightarrow \frac{4(2n - 1)}{3} = \frac{44}{3}\]
\[ \Rightarrow (2n - 1) = 11\]
\[ \Rightarrow 2n = 12\]
\[ \Rightarrow n = 6\]
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