Question

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

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

Answer 1

Answer: (B)

We know ${}^n{P}_r={}^n{C}_r(r)!$
$\Rightarrow \frac{{}^n{P}_r}{r!}={}^n{C}_r$
Take ${\sum }_{r=1}^n$ on both sides, we get
${\sum }_{r=1}^n\frac{{}^n{P}^r}{r!}={\sum }_{r=1}^n{}^n{C}_r$
${=}^n{C}_1{+}^n{C}_2{+}^n{C}_3+.....{+}^n{C}_n$
$=\left({}^n{C}_0{+}^n{C}_1{+}^n{C}_2+....{+}^n{C}_n\right)-1$
$=2^n-1$