First slide

Permutations

Question

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

Moderate

A

10, 11
B

9, 10
C

8, 7
D

6,7

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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