If 5 P(4,n)= 6 P (5,n-1), find n.

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Solution

We have,
5P (4,n)= 6, (5,n-1)
$\Rightarrow 5 \times \frac{1}{(4 - n)!} = 6 \times \frac{5!}{[5 - (n - 1)]!}$
$[∵ n_{p_{r}} = \frac{n!}{(n - r)!}]$
$\Rightarrow 5 \times \frac{4!}{(4 - n)!} = \frac{6 \times 5 \times 4!}{[5 - n + 1]!}$
$\Rightarrow \frac{1}{(4 - n)!} = \frac{6}{(6 - n) (6 - n - 1) (6 - n - 2)!}$
$\Rightarrow \frac{1}{(4 - n)!} = \frac{6}{(6 - n) (5 - n) (4 - n)!}$
$\Rightarrow \frac{(6 - n) (5 - n) (4 - n)!}{(4 - n)!} = 6$
$\Rightarrow (6 - n) (5 - n) = 6$
$\Rightarrow 30 - 6 n - 5 n + n^{2} = 6$
$\Rightarrow n^{2} - 11 n + 30 = 6$
$\Rightarrow n^{2} - 11 n + 24 = 0$
$\Rightarrow n^{2} - 8 n - 3 n + 24 = 0$
$\Rightarrow n (n - 8) - 3 n + 24 = 0$
$\Rightarrow (n - 8) (n - 3) = 0$
$\Rightarrow n - 3 = 0$
$[\begin{matrix} ∵ n \leq 4 ∴ n \neq 8 \end{matrix}]$
$\Rightarrow n = 3$
Hence, n=3

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