Question

The highest power of $18$ contained in ${}^{50}{C}_{25}$.

• A. $3$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

${{}^{50}C}_{25}$ = $\frac{50!}{(25!)(25!)}$

$18=3\times 2\times 2$

We have to find numbers of $2^{\prime }$s and $3^{\prime }$s in $50!$ and $25!$

$2^{\prime }$s in $50!:$

$=\left[\frac{50}{2}\right]+\left[\frac{50}{4}\right]+\left[\frac{50}{8}\right]+\left[\frac{50}{16}\right]+\left[\frac{50}{32}\right]$

$=25+12+6+3+1$

$=47$

$3^{\prime }$s in $50!$

$=\left[\frac{50}{3}\right]+\left[\frac{50}{9}\right]+\left[\frac{50}{27}\right]$

$=16+5+1$

$=22$

Similarly no. of $2^{\prime }$s in $25=22$ and no. of 3's in $25=10$

$\therefore$ In $\frac{50!}{(25!)(25!)}\to$ no. of $2^{\prime }s=47-44=3$ and no. of $3^{\prime }s=22-20=2.$

So, only ${18}^1$ is possible.

Hence the answer is $1.$