JEE Mains · Maths · STD 11 - 6. permutation and combination
If \({ }^{2 n+1} P_{n-1}:{ }^{2 n-1} P_n=11: 21\), then \(n^2+n+15\) is equal to:
- A \(44\)
- B \(43\)
- C \(42\)
- D \(45\)
Answer & Solution
Correct Answer
(D) \(45\)
Step-by-step Solution
Detailed explanation
\(\frac{(2 n +1) !( n -1) !}{( n +2) !(2 n -1) !}=\frac{11}{21}\) \(\Rightarrow \frac{(2 n +1)(2 n )}{( n +2)( n +1) n }=\frac{11}{21}\) \(\Rightarrow \frac{2 n +1}{( n +1)( n +2)}=\frac{11}{42}\) \(\Rightarrow n =5\) \(\Rightarrow n ^2+ n +15=25+5+15=45\)
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