Question

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

Answer 1

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 $\left[\right.\star\left]\right.$ 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 $\left[\right.\star\left]\right.$ 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 $^{100}C_{50}$
$=24-11$
$=13$