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)

$2^n - 1$

Answer 2

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