A number theory problem by Jai sahota

$C(200,100)= \binom{200}{100} =\frac{200!}{100!100!}$ Total numbers of trailing zeros hold by $\binom{200}{100}$ are the prime factor of $5$ . So highest power of prime $p= 5$ for $200!$ , $100!$ is given by \begin{aligned}& H_p(n!) & =\displaystyle\sum_{k=1}^{m}\left\lfloor\frac{n}{p^k}\right\rfloor\phantom{cccccccc}\text{where p^m≤n<p^{m+1}} \end{aligned} $\begin{aligned} H_5(200!) & = \left\lfloor\frac{200}{5}\right\rfloor +\left\lfloor\frac{200}{25}\right\rfloor+\left\lfloor\frac{200}{125}\right\rfloor \\& = 40+9+1 = 49\end{aligned}$ $\begin{aligned} H_5(100!) & = \left\lfloor\frac{100}{5}\right\rfloor +\left\lfloor\frac{100}{25}\right\rfloor \\& = 20+4= 24\end{aligned}$

It can be noted that the total number of zeros in denominator are $2\times24 = 48$ .There are altogether $49$ zeros in the numerator and $48$ zeros in denominator. Remaining numbers of are ending zeros of $\binom{200}{100}$ ie $1$ .

$\binom{200}{100}= \frac{200!}{100! \times 100!}$ ,which again is equal to $\large \frac{200 \times 199 \times 198 \times \cdots \times 102 \times 101}{100 \times 99 \times 98 \times \cdots \times 2 \times 1}$ Now by using floor function\greatest integer function on $100!$ we will get number of $0$ 's equal to $24$ and similarly using floor function on $200!$ , we will get number of $0$ 's equal to $49$ .But the product in the numerator will have only $49-24=25$ $0$ 's. Hence, the number of zeroes in the expression will be $\boxed{1}$ because $24$ $0$ 's will get cancelled from numerator and denominator both.