Question

The number of zero(s) in ${}^{500}{C}_{20}$ is

Answer 1

We know, number of zeros in $n!=$ maximum exponent of $5$ in $n!$

${}^{500}{C}_{20}=\frac{500!}{20!\cdot 480!}$
Now,
$E_5(500!)=\left[\frac{500}{5}\right]+\left[\frac{500}{25}\right]+\left[\frac{500}{125}\right]+\cdots$
$=100+20+4+0+\cdots$
$=124$
$E_5(20!)=\left[\frac{20}{5}\right]+\cdots$
$=4+0+\cdots =4$
$E_5(480!)=\left[\frac{480}{5}\right]+\left[\frac{480}{25}\right]+\left[\frac{480}{125}\right]+\cdots$
$=96+19+3+0+\cdots$
$=118$
$\Rightarrow$ Maximum exponent of $5$ in ${}^{500}{C}_{20}=124-(118+4)=2$
$\therefore$ Number of zeros in ${}^{500}{C}_{20}=2$