aefenglommung (aefenglommung) wrote,
aefenglommung
aefenglommung

They must be Democrats

The problem given was, there are eleven members of a political party organization: 3 black males, 2 black females, 2 white males, and 4 white females. Party rules dictate that of the 3 officers, one must be black and one must be female. How many possible combinations of the members yield that result?

Well, I don't know the formula, but here's my work.

Call the 3 black men A, B, and C.
Call the 2 black women D and E.
Call the 2 white men F and G.
Call the 4 white women H, I, J, and K.

In order for a white male to be an officer, one of the other officers must be black. If the second officer is a black male, then the third officer must be a female; if the second officer is a black female, then the third officer can be any remaining member of the committee.

F or G + A requires D, E, H, I, J, or K = 12 combinations
F or G + B requires D, E, H, I, J, or K = 12 combinations
F or G + C requires D, E, H, I, J, or K = 12 combinations
F or G + D requires E, F or G, H, I, J, or K = 12 combinations
F or G + E requires F or G, H, I, J, or K = 10 combinations
Total combinations allowing for one or two white males = 58 combinations

Assuming that no white male is an officer, we start with one officer being a black male. The second officer must then be female. The third officer can be anyone except a white male (whose combinations are all used up in the first section).

A + D requires B, C, E, H, I, J, or K = 7 combinations
A + E requires B, C, H, I, J, or K = 6 combinations
A + H requires B, C, I, J, or K = 5 combinations
A + I requires B, C, J, or K = 4 combinations
A + J requires B, C, or K = 3 combinations
A + K requires B or C = 2 combinations
Total combinations allowing for black male A = 27 combinations
Substituting B for A will yield the same results, less 6 combinations already used = 21 combos
Substituting C or A will yield the same results, less 6 combinations already used = 21 combos
Total combinations allowing a mix of black males and white or black females = 69 combinations

Assuming an all-female officer cadre yields the following combinations.

D or E (1 black female) requires 2 white females in combos H and I, H and J, H and K, I and J, I and K, or J and K = 12 combinations
D + E (2 black females) requires 1 white female H, I, J, or K = 4 combinations
Total combinations for all-female officer cadres = 16 combinations

58 combos involving white males + 69 combos involving black males and white or black females + 16 combinations involving black females and white females = 143 possible combinations
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