Prove that $^{n + 1} P_{r} = \frac{(n + 1)}{(n - r + 1)} \times^{n} P_{r}$ $""^(n+1)P_r=((n+1))/((n-r+1))xx""^nP_r$ ?

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Answer 1 · 2016-09-14T05:18:05

Please see below.

The formula for the number of possible permutations of $r$ $r$ objects from a set of $n$ $n$ is written as $^{n} P_{r}$ $""^nP_r$ and is

$^{n} P_{r} = \frac{n!}{(n - r)!}$ $""^nP_r=(n!)/((n-r)!)$ ,

where $k!$ $k!$ is defined as $k! =kxx(k-1)xx(k-2)...3xx2xx1$

Hence $""^(n+1)P_r=((n+1)!)/((n+1-r)!)=((n+1)!)/((n-r+1)!)$

= $((n+1)xxn!)/((n-r+1)xx(n-r)!)$

= $((n+1))/((n-r+1))xx(n!)/((n-r)!)$

= $((n+1))/((n-r+1))xx""^nP_r$

Prove that ""^(n+1)P_r=((n+1))/((n-r+1))xx""^nP_rn+1Pr=(n+1)(n−r+1)×nPr""^(n+1)P_r=((n+1))/((n-r+1))xx""^nP_r?