Let $\mathrm{E}=\left[\frac{1}{3}+\frac{1}{50}\right]+\left[\frac{1}{3}+\frac{2}{50}\right]+\left[\frac{1}{3}+\frac{3}{50}\right]+\dots +$ up to 50 terms, then the exponent of 2 in E! is

Let, $\mathrm{E}=\left[\frac{1}{3}+\frac{\mathrm{x}}{50}\right]$ where x = 1,2, ... ,50 

For, $1\le \mathrm{x}\le 33,\frac{1}{3}<\frac{1}{3}+\frac{\mathrm{x}}{50}<1$

$\text{∴}\left[\frac{1}{3}+\frac{\mathrm{x}}{50}\right]=0, \mathrm{for} 1\le \mathrm{x}\le 33$

For $34\le \mathrm{x}\le 50,1<\frac{1}{3}+\frac{\mathrm{x}}{50}<\frac{4}{3}$

$\Rightarrow \left[\frac{1}{3}+\frac{x}{50}\right]=1$ , For $34\le \mathrm{x}\le 50$

Thus, E=17 and p=2

$\begin{array}{lc}\because & 2^4<17<2^5\\ \therefore & \mathrm{a}=4\end{array}$

Exponent of 2 in $\left(17\right)!=\left[\frac{17}{2}\right]+\left[\frac{17}{4}\right]+\left[\frac{17}{8}\right]+\left[\frac{17}{16}\right]$

                                   $=8+4+2+1=15$