Question

The highest power of $18$ contained in ${}^{100}{C}_{50}$ is

Answer 1

Answer: 13

Given, ${}^{100}{C}_{50}=\frac{100!}{50!50!}=\frac{100!}{{\left(50!\right)}^2}$
Now exponent of $2$ in $100!$
$=\left[\frac{100}{2}\right]+\left[\frac{100}{2^2}\right]+\left[\frac{100}{2^3}\right]+\left[\frac{100}{2^4}\right]+\left[\frac{100}{2^5}\right]+\left[\frac{100}{2^6}\right]$
$=\left[50\right]+\left[25\right]+\left[12.5\right]+\left[6.2\right]+\left[3.1\right]+\left[1.5\right]$
(Here $[\star ]$ denotes $G.I.F.$ )
$=50+25+12+6+3+1$
$=97$
Now exponent of $3$ in $100!$
$=\left[\frac{100}{3}\right]+\left[\frac{100}{3^2}\right]+\left[\frac{100}{3^3}\right]+\left[\frac{100}{3^4}\right]$
$=\left[33.3\right]+\left[11.1\right]+\left[3.70\right]+\left[1.2\right]$
$=33+11+3+1;$ Here $[\star ]$ denotes $G.I.F.$
$=48$
Thus exponent of $18$ in $100!=24$
Now exponent of $2$ in $50!$
$=\left[\frac{50}{2}\right]+\left[\frac{50}{2^2}\right]+\left[\frac{50}{2^3}\right]+\left[\frac{50}{2^4}\right]+\left[\frac{50}{2^5}\right]$
$=47$
Now exponent of $18$ in $50!=11$
Hence, highest power of $18$ in ${}^{100}{C}_{50}$
$=24-11$
$=13$