If p(11,r)= P (12, r-1) find r.

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Solution

We have,
P (11,r) = P(12,r-1)
$\Rightarrow \frac{11!}{(11 - r)!} = \frac{12!}{[12 - (r - 1)]!} [∵ n_{p_{r}} = \frac{n!}{(n_{r})!}] \Rightarrow \frac{11!}{(11 - r)!} = \frac{12 \times 11!}{[12 - r + 1]!} \Rightarrow \frac{1!}{(11 - r)!} = \frac{12}{[13 - r]!} \Rightarrow \frac{1!}{(11 - r)!} = \frac{12}{(13 - r) (12 - r) (11 - r)!} = 12 \Rightarrow (13 - r) (12 - r) = 12 \Rightarrow 156 - 13 r - 12 r + r^{2} = 12 \Rightarrow r^{2} - 25 r + 156 - 12 = 0 \Rightarrow r^{2} - 25 r + 144 = 0 \Rightarrow r^{2} - 16 r - 9 r + 144 = 0 \Rightarrow r (r - 16) - 9 (r - 16) = 0 \Rightarrow r - 9 = 0$
$[\begin{matrix} ∵ r \leq 11 ∴ r \neq 16 \end{matrix}]$
$\Rightarrow r = 9$

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