Question

The value of $20! + \frac{21!}{1!} + \frac{22!}{2!} + ..... + \frac{60!}{40!}$ is

• A. $20! {^{61}C_{20}}$
• B. $21! {^{60}C_{20}}$
• C. $20! {^{61}C_{21}}$
• D. $21! {^{60}C_{19}}$

Answer 1

Answer: (C)

$20! {^{61}C_{21}}$

Answer 2

We have,
$20! +\frac{21!}{1!} + \frac{22!}{2!} + ... +\frac{60!}{40!}$
$=20! \left[ 1+\frac{21!}{20!1!} + \frac{22!}{20!2!} +... +\frac{60!}{40!20!}\right]$
$= 20! \left[\,^{21}C_{0} +\,^{21}C_{1} +\,^{22}C_{2} +...+ \,^{60}C_{40}\right]$
$= 20!\left[\,^{22}C_{1} + \,^{22}C_{2}+...+\,^{60}C_{40}\right]$
$\left[\,^{n}C_{r-1} +\,^{n}C_{r} = \,^{n+1}C_{r}\right]$
$= 20! \left[ \,^{61}C_{40}\right] = 20! \cdot \,^{61}C_{21}$