Prove with or without the use of the formula.

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

Open in App

Solution

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

RHS = $n \frac{(n - 1)!}{(n - r)!} = \frac{n!}{(n - r)!} =^{n} P_{r}$ = LHS

Suggest Corrections

Similar questions

^{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}

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

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

\frac{^{n} P_{r}}{^{n} P_{r - 2}} = (n - r + 2) (n - r + 1)

Join BYJU'S Learning Program