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If ( 2 N ) ! 3 ! ( 2 N − 3 ) ! and N ! 2 ! ( N − 2 ) ! Are in the Ratio 44 : 3, Find N. - Mathematics

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प्रश्न

If \[\frac{(2n)!}{3! (2n - 3)!}\]  and \[\frac{n!}{2! (n - 2)!}\]  are in the ratio 44 : 3, find n.

 

 

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उत्तर

\[\frac{(2n)!}{3!(2n - 3)!}: \frac{n!}{2!(n - 2)!} = 44: 3 \]
\[ \Rightarrow \frac{(2n)!}{3!(2n - 3)!} \times \frac{2!(n - 2)!}{n!} = \frac{44}{3}\]
\[ \Rightarrow \frac{(2n)(2n - 1)(2n - 2) [(2n - 3)!]}{3(2!)(2n - 3)!} \times \frac{2!(n - 2)!}{n(n - 1) [(n - 2)!]} = \frac{44}{3}\]
\[ \Rightarrow \frac{(2n)(2n - 1)(2n - 2)}{3} \times \frac{1}{n(n - 1)} = \frac{44}{3}\]
\[ \Rightarrow \frac{(2n)(2n - 1)(2)(n - 1)}{3} \times \frac{1}{n(n - 1)} = \frac{44}{3}\]
\[ \Rightarrow \frac{4(2n - 1)n(n - 1)}{3} \times \frac{1}{n(n - 1)} = \frac{44}{3}\]
\[ \Rightarrow \frac{4(2n - 1)}{3} = \frac{44}{3}\]
\[ \Rightarrow (2n - 1) = 11\]
\[ \Rightarrow 2n = 12\]
\[ \Rightarrow n = 6\]

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Factorial N (N!) Permutations and Combinations
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अध्याय 16: Permutations - Exercise 16.1 [पृष्ठ ५]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 16 Permutations
Exercise 16.1 | Q 10 | पृष्ठ ५

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