Question

Show that $\rm (n+1)\ ^{n}P_{r} = \rm (n-r+1)\ ^{(n+1)}P_{r}$.

Answer 1

Answer:

Step-by-step explanation:

Results used:

1.$nP_r=\frac{n!}{(n-r)!}$

2. n! = (n-1)! n

$(n+1)\:nP_r\\\\=(n+1)\frac{n!}{(n-r)!}\\\\=\frac{n!(n+1)}{(n-r)!}$

Multiply both numerator and denominator

by (n-r+1)

$=(n-r+1)\frac{n!(n+1)}{(n-r)!(n-r+1)}\\\\=(n-r+1)\frac{n!(n+1)}{(n-r)!(n+1-r)}\\\\=(n-r+1)\frac{(n+1)!}{(n+1-r)!}\\\\=(n-r+1)\:(n+1)P_r$