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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 number of zeros at the end of $108 ! $ is -

Permutations and Combinations

Solution:

Product of 5's & 2's constitute 0's at the end of a number
$\Rightarrow$ No. of 0 's in $108 ! $ = exponent of 5 in $ 108 !$
(Note that exponent of 2 will be more than exponent of 5 in $ 108 !)$
$\Rightarrow\left[\frac{108}{5}\right]+\left[\frac{21}{5}\right]=21+4=25$