How to find exponent of a number in a combination? How do I find the exponent of $7$ in $^{100}C_{50}$ that is, $\dfrac{100!}{(100-50)!\cdot 50!} =\dfrac{100!}{50!\cdot 50!}$,
this question was out of the blue, and I haven't been able to find any worked out examples on it. I'm new to permutations and combinations and frankly didn't understand what the question is asking. :-(
It'll be great if someone could explain the question and also suggest the way to go about it..  
cheers! :-)
 A: Take a prime $p$ and an integer $n$. By Legendre's theorem the largest power of $p$ dividing $n!$ is given by the sum:
$$
\sum_{k \ge 1} \left\lfloor \frac{n}{p^k} \right\rfloor
$$
(the sum is really finite). Use this thrice.
A: Note that there are $14$ multiples of $7$ less than or equal to $100$, and $2$ of these are multiples of $49 = 7^2$ (and $0$ multiples of $343 = 7^3$, etc.).  So all in all, $7^{16}$ (counting each of the multiples of $7$ once, and then counting the multiples of $49$ an additional time, so $14 + 2 = 16$) is the largest exponent of $7$ that divides $100!$.
Repeating this for $50!$, we get that $7^8$ is the largest exponent of $7$ dividing $50!$.  So $7^{16} / 7^8 7^8 = 1$.
$7$ does not divide $\left( \begin{array}{c} 100 \\ 50 \end{array} \right)$.
A: In this answer, it is shown that for a prime $p$, the number of factors of $p$ that divide $\binom{n}{k}$ is
$$
\frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}\tag{1}
$$
where $\sigma_p(n)$ is the sum of the digits of $n$ in base-$p$.
$100_\text{ten}=202_\text{seven}$ and $50_\text{ten}=101_\text{seven}$. Therefore, the number of factors of $7$ in $\binom{100}{50}$ is
$$
\frac{2+2-4}{6}=0\tag{2}
$$
Thus, $\binom{100}{50}$ is not divisible by $7$.
