Question:
The value of $\sum^{n}_{r =1} \frac{^nP_r}{r!} $ is :
The value of $\sum^{n}_{r =1} \frac{^nP_r}{r!} $ is :
Updated On: Jul 7, 2022
- $2^n$
- $2^n - 1 $
- $2^{n - 1} $
- $2^n + 1 $
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The Correct Option is B
Solution and Explanation
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 $
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Concepts Used:
Permutations
A permutation is an arrangement of multiple objects in a particular order taken a few or all at a time. The formula for permutation is as follows:
\(^nP_r = \frac{n!}{(n-r)!}\)
nPr = permutation
n = total number of objects
r = number of objects selected
Types of Permutation
- Permutation of n different things where repeating is not allowed
- Permutation of n different things where repeating is allowed
- Permutation of similar kinds or duplicate objects