Permutation Combination Question 17
Question 17 - 31 January - Shift 2
$(6 A+11)^{2}=125$
If ${ }^{2 n+1} P _{n-1}:{ }^{2 n-1} P_n=11: 21$, then $n^{2}+n+15$ is equal to ___________
Show Answer
Answer: 45
Solution:
Formula: Permutation formula
$\frac{(2 n+1) !(n-1) !}{(n+2) !(2 n-1) !}=\frac{11}{21}$
$\Rightarrow \frac{(2 n+1)(2 n)}{(n+2)(n+1) n}=\frac{11}{21}$
$\Rightarrow \frac{2 n+1}{(n+1)(n+2)}=\frac{11}{42}$
$\Rightarrow n=5$
$\Rightarrow n^{2}+n+15=25+5+15=45$
