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Q. Let $p$ be a prime number and $n$ be a positive integer, then exponent of $p$ is $n !$ is denoted by $E_p(n !)$ and is given by
$E_p(n !)=\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\left[\frac{n}{p^3}\right]+\ldots .+\left[\frac{n}{p^k}\right] $
where $ p ^{ k }< n < p ^{ k +1}$
and $[ x ]$ denotes the integral part of $x$.
If we isolate the power of each prime contained in any number $N$, then $N$ can be written as
where $\alpha_i$ are whole numbers.
The exponent of $12$ in $100 !$ is -

Permutations and Combinations

Solution:

As $12=2^2 .3$, here we have to calculate exponent of $2$ and exponent of $3$ in $100 !$ exponent of $2$
$=\left[\frac{100}{2}\right]+\left[\frac{50}{2}\right]+\left[\frac{25}{2}\right]+\left[\frac{12}{2}\right]+\left[\frac{6}{2}\right]+\left[\frac{3}{2}\right]=97$
exponent of $3=\left[\frac{100}{3}\right]+\left[\frac{33}{3}\right]+\left[\frac{11}{3}\right]+\left[\frac{3}{3}\right]=48$
Now, $12=2 2 3$
we require two $2's$ & one $3$
$\therefore $ exponent of $3$ will give us the exponent of $12$ in $100 !$ i.e. $48$