I was given this hint in a different problem,
Now use that a prime $p$ occurs in $n!$ with multiplicity exactly $\lfloor n/p\rfloor + \lfloor n/p^2\rfloor + \lfloor n/p^3\rfloor + \lfloor n/p^4\rfloor +\ldots$
For $$P=\frac{200!}{2^{100}\cdot 100!}$$ And the prime being $p = 3$.
How is the claim for $n!$ true?
Consider $n \ge 6$.
$$n = 6 \implies n! = 720$$
$$720/3 = 240 \to 240/3 = 80 \implies \text{3 comes in twice.}$$
Then, $[6/3] + [6/9] = 2$.
So suppose $3$ occurs in $n!$ with multiplicity, $\lfloor n/p\rfloor + \lfloor n/p^2\rfloor + \lfloor n/p^3\rfloor + \lfloor n/p^4\rfloor +\ldots$
It is required to show that, for $(n+1)!$, $3$ occurs in multiplicity,
$\lfloor n+1/p\rfloor + \lfloor n+1/p^2\rfloor + \lfloor n+1/p^3\rfloor + \lfloor n+1/p^4\rfloor +\ldots$
$(n+1)! = (n+1)n!$.
But I cant prove anything else.
Even intuitively, why does this make sense?
