8. We arrange several "0"s and "1"s in a row, defining the capacity of this arrangement as the number of times "10" appears: for example, the capacity of "1010" is 2, and the capacity of "0101" is 1. If we arrange 50 "0"s and 50 "1"s in a row, we will get $\binom{100}{50}$ different arrangements. Then the sum of the capacities of these arrangements is $\qquad$
Consider the occurrence of "10" at the $i, i+1$ positions. Since the remaining 98 positions have 49 "0"s and 49 "1"s, the number of corresponding permutations is $C_{98}^{49}$, meaning each "10" is counted $C_{98}^{49}$ times. Therefore, the total sum of the capacities of all permutations is $99 C_{98}^{49}$.