If ${ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_{n}=11: 21$,
then $n^{2}+n+15$ is equal to :
Answer (integer)
45
Solution
$\frac{\frac{(2 n+1) !}{(n+2) !}}{\frac{(2 n-1) !}{(n-1) !}}=\frac{11}{21}$
<br/><br/>$\frac{(2 n+1) 2 n}{(n+2)(n+1) n}=\frac{11}{21}$
<br/><br/>$84 n+42=11\left(n^{2}+3 n+2\right)$
<br/><br/>$11 n^{2}-51 n-20=0$
<br/><br/>$(n-5)(11 n+4)=0$
<br/><br/>$n=5, \frac{-4}{11}$ (Rejected $)$
<br/><br/>$n^{2}+n+15=45$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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