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NDA 2011
Exam
NDA
Year
2011
Subject
Mathematics
Chapter
Permutation and Combination

NDA 2011 · Mathematics · Permutation and Combination

What is the value of n , if P(15, n-1): P(16, n-2)=3: 4 ?

What is the value of nn, if P(15,n−1):P(16,n−2)=3:4P(15, n-1): P(16, n-2)=3: 4 ?

Options

  1. A10
  2. B12
  3. C14
  4. D15

Correct answer

C. 14

Step-by-step solution

P(15,n−1):P(16,n−2)=3:4⇒15Pn−116Pn−2=34\begin{aligned} & \mathrm{P}(15, n-1): \mathrm{P}(16, n-2)=3: 4 \\ & \Rightarrow \quad \frac{15 \mathrm{P}_{n-1}}{16 \mathrm{P}_{n-2}}=\frac{3}{4}\end{aligned}\Rightarrow \quad \frac{\frac{15!}{15-x+1!}}{\frac{16!}{(16-n+2)}}=\frac{3}{4}\Rightarrow \quad \frac{15!18-n!}{16!16-n!}=\frac{3}{4}(8−n)(17−n)16=34n2−35n+294=0n2−21n−14n+294=0n(n−21)−14(n−21)=0\begin{aligned} & \frac{(8-n)(17-n)}{16}=\frac{3}{4} \\ & n^2-35 n+294=0 \\ & n^2-21 n-14 n+294=0 \\ & n(n-21)-14(n-21)=0\end{aligned}\Rightarrow \quad n=14,21$.
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