This seemed like a very complicated problem at first . . .
It was much simpler than I thought . . .Count the ways that two 50-length binary strings can be formed
so that in total they contain ten 1's and ninety 0's.
and in the second string. all the 0's must precede all the 1's (if any).
Consider the first group.
It contains 50 digits: some 0's, some 1's.
Suppose the first group has seven 1's and forty-three 0's.
The first group can be any of: . possible strings.
But the second group is already determined!
It will have three 1's and forty-seven 0's
So we are concerned with the number of possible first groups.
The answer is: .
You can crank it out . . . I'll wait in the car.