Zero - Zero is a big score.

$^{125}C_{62}=\frac{125!}{(62!)(63!)}$ Number of trailing 0s $=$ how many factors of 10 we have in this expression. Since $10=2×5$ , and the number of factors of 2 greatly exceed that of 5, all we need to do is count how many factors of 5 we have in the expression:

For 125!:

There are $\frac{125}{5}=25$ numbers with 1 factor of 5. But $25=5^2, 50=2×5^2, 75=3×5^2$ $100=4×5^2,$ and $125=5^3$ so we have an additional 6 factors of 5, hence $10^{31}||125!$

For 63! and 62!:

There are equal numbers of factors of 5 in $62!$ and $63!$ as $63!=62!×63$ and 63 has no factors of 5. There are $\frac{60}{5}=12$ numbers in each 62! and 63! with 1 factor of 5. In addition, 25 and 50 have an extra factor of 5 as shown earlier. Hence $62!×63!$ has $14×2=28$ factors of 5, i.e $5^{28}||(62!×63!)$

So in total, $^{125}C_{62}$ has $\frac{5^{31}}{5^{28}}=5^3$ , hence there are 3 zeroes trailing in this expression.