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A play has two male roles, two female roles, and 2 roles that can be either gender.
Only a man can be assigned to a male role, and only a woman can be assigned to a female role.
If five men and six women audition, in how many ways can the six roles be assigned?
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The roles are M1, M2 (played by men, only); F1, F2 (played by women, only), and A1, A2 (played by any gender).
There are 5 choices for role M1; 4 choices for role M2; and after that 5-2 = 3 men are in reserve.
There are 6 choices for role F1; 5 choices for role F2; and after that 6-2 = 4 women are in reserve.
So, now we have 3+4 = 7 persons in the reserve for roles A1 and A2.
The total number of possible combinations is (5*4) * (6*5) * (7*6) = 25200. ANSWER
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Solved and carefully explained ( but not pedantic ).
"Pedantic" always means "boring".
Notice that, although I used the term "combinations" in my post,
the order inside these combinations DOES MATTER ( ! )
So actually, these "combinations" are, in fact, "permutations" with some imposed restrictions.
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