If n+5pn+1=12(11)(n−1) n+3pn then n is equal to

12(11)(n−1)= n+5Pn+1 n+3Pn=(n+5)!4!×3!(n+3)!⇒ 12(11)(n−1)=14(n+5)(n+4)⇒ 22n−22=n2+9n+20⇒ n2−13n+42=0⇒n=6,7

If n+5pn+1=12(11)(n−1) n+3pn then n is equal to

12(11)(n−1)= n+5Pn+1 n+3Pn=(n+5)!4!×3!(n+3)!⇒ 12(11)(n−1)=14(n+5)(n+4)⇒ 22n−22=n2+9n+20⇒ n2−13n+42=0⇒n=6,7

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If $^{n + 5} p_{n + 1} = \frac{1}{2} (11) (n - 1) (^{n + 3} p_{n})$ then $n$ is equal to

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6,7

10, 11

9, 10

8, 7

answer is D.

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Detailed Solution

$\frac{1}{2} (11) (n - 1) = \frac{^{n + 5} P_{n + 1}}{^{n + 3} P_{n}}$

$= \frac{(n + 5)!}{4!} \times \frac{3!}{(n + 3)!}$

$\Rightarrow \frac{1}{2} (11) (n - 1) = \frac{1}{4} (n + 5) (n + 4)$

$\begin{array}{ll} \Rightarrow 22 n - 22 = n^{2} + 9 n + 20 \\ \Rightarrow n^{2} - 13 n + 42 = 0 \Rightarrow n = 6, 7 \end{array}$

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If $^{n + 5} p_{n + 1} = \frac{1}{2} (11) (n - 1) (^{n + 3} p_{n})$ then $n$ is equal to