First we look at the probability of producing a total ensemble which is 1/(n+m)!. This is in any random order and any part either m or n
To further see what is going on let’s take some example let’s have 6 n spoilers and 3 m spoilers:
n_1,n_2,n_3,n_4,n_5,n_6m_1,m_2,m_3
Let’s take any random order:
n_4,n_1, m_3,m_2,m_1, n_2,n_1,n_,6,n_5,n_3
As it can seen we can treat the sequence of m spoiler as one called M and within M there are m! possible combinations
so now we can take another sample but now we look at the overall ensemble
n_3,n_1,n_6,n_4, M, n_2,n_5
So it can observed that we can combine this in (6+1)! Ways therefore for n spoilers we could combine them (n+1)! Ways of combining the M sequence. So the total probability will be:
(n+1)!m!/(m+n)!