Question

The value of $\sum^{n}_{r =1} \frac{^nP_r}{r!} $ is :

• A. $2^n$
• B. $2^n - 1 $
• C. $2^{n - 1} $
• D. $2^n + 1 $

Answer 1

Answer: (B)

We know ${^nP_r} = {^nC_r} (r) !$
$\Rightarrow \, \frac{^nP_r}{r!} = {^nC_r}$
Take $\sum^n_{r = 1}$ on both sides, we get
$\sum^{n}_{r=1} \frac{^{n}P^{r}}{r!} = \sum^{n}_{r=1} {^{n}C_{r}} $
$= ^{n}C_{1} + ^{n}C_{2} + ^{n}C_{3} + ..... +^{n}C_{n} $
$= \left(^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2} + ....+^{n}C_{n}\right) - 1 $
$= 2^{n} - 1 $