What is the value of $n.P(n-1, r-1)$ ?

P(n, r)

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P(n, n)

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P(r, r)

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Solution

The correct option is A $P(n, r)$
$n.P(n-1, r-1) = n \times \frac{(n - 1)!}{[(n - 1) - (r - 1)]!}$

$\Rightarrow n.P(n-1, r-1) = \frac{n . (n - 1) . (n - 2) . . . . . . .3 .2 .1}{[n - 1 - r + 1]}$

We know that $1.2.3.4...... (n - 1) . n = n!$

$\Rightarrow n.P(n-1, r-1) = \frac{n!}{(n - r)} = P(n, r)$

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