Question:
The value of $20! + \frac{21!}{1!} + \frac{22!}{2!} + ..... + \frac{60!}{40!} $ is
The value of $20! + \frac{21!}{1!} + \frac{22!}{2!} + ..... + \frac{60!}{40!} $ is
Updated On: Jun 20, 2022
- $20! {^{61}C_{20}}$
- $21! {^{60}C_{20}}$
- $20! {^{61}C_{21}}$
- $21! {^{60}C_{19}}$
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The Correct Option is C
Solution and Explanation
We have,
$20! +\frac{21!}{1!} + \frac{22!}{2!} + ... +\frac{60!}{40!}$
$ =20! \left[ 1+\frac{21!}{20!1!} + \frac{22!}{20!2!} +... +\frac{60!}{40!20!}\right]$
$= 20! \left[\,^{21}C_{0} +\,^{21}C_{1} +\,^{22}C_{2} +...+ \,^{60}C_{40}\right] $
$ = 20!\left[\,^{22}C_{1} + \,^{22}C_{2}+...+\,^{60}C_{40}\right]$
$ \left[\,^{n}C_{r-1} +\,^{n}C_{r} = \,^{n+1}C_{r}\right] $
$= 20! \left[ \,^{61}C_{40}\right] = 20! \cdot \,^{61}C_{21} $
$20! +\frac{21!}{1!} + \frac{22!}{2!} + ... +\frac{60!}{40!}$
$ =20! \left[ 1+\frac{21!}{20!1!} + \frac{22!}{20!2!} +... +\frac{60!}{40!20!}\right]$
$= 20! \left[\,^{21}C_{0} +\,^{21}C_{1} +\,^{22}C_{2} +...+ \,^{60}C_{40}\right] $
$ = 20!\left[\,^{22}C_{1} + \,^{22}C_{2}+...+\,^{60}C_{40}\right]$
$ \left[\,^{n}C_{r-1} +\,^{n}C_{r} = \,^{n+1}C_{r}\right] $
$= 20! \left[ \,^{61}C_{40}\right] = 20! \cdot \,^{61}C_{21} $
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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:
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nPr = permutation
n = total number of objects
r = number of objects selected
Types of Permutation
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