Prove that $^{n} P_{r} =^{n - 1} P_{r} + r .^{n - 1} P_{r - 1}$ .

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Solution

L.H.S. = $^{n} P_{r} = \frac{n!}{(n - r)!}$ ..(1)
R.H.S. = $^{n - 1} P_{r} + r .^{n - 1} P_{r - 1} = \frac{(n - 1)!}{(n - r - 1)!} + \frac{(n - 1)!}{(n - r)!}$
= $\frac{(n - 1)!}{(n - r - 1)!} [1 + \frac{r}{n - r}]$
= $\frac{n . (n - 1)!}{(n - 1) . (n - r - 1)!} = \frac{n!}{(n - r)!}$ ..(2)
Hence from (1) and (2)

$^{n} P_{r} =^{n - 1} P_{r} + r .^{n - 1} P_{r - 1}$ .

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^{n} P_{r} =^{n - 1} P_{r} + r .^{n - 1} P_{r - 1}

^{n} P_{r} =^{(n - 1)} P_{r} + r^{(n - 1)} P_{(r - 1)}

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