A Mathematics department consists of 5 female and 5 male teachers.
How many committees of 3 teachers can be chosen which contain at least one female and at least one male? a) 100 b) 120 c) 200 d) 2500
So I found several ways of doing this question Method 1: 10C3 - 5C3 - 5C3 = 100 (total number of combinations - number of combinations with no male - number of combination with no female)
Method 2: 2x(5C1 x 5C2) = 100 [number of combinations with 1 male and 2 female multiplied by 2 (as it can be 1 female and 1 male)]
These two both work to give the correct answer but the following method was my initial attempt.
My Method: 5C1 x 5C1 x 8C1 = 200 [The number of combinations of choosing 1 from male x number of combinations of choosing 1 from female x number of ways you can choose 1 from the remaining 8 people]
Can you please explain why my method does not work?