Six Canadian and 6 Mexican delegates attend an international trade conference. A three-delegate committee will be selected from the 12 delegates. How many three-delegate committees can be formed if each possible committee must include at least one Mexican delegate?

A) 260

B) 240

C) 220

D) 200

E) 20

HINT: Subtract from total possible groups those that do not include any Mexican delegate.

ReplyDeleteSo you are saying 12 pick 3 minus 6 pick 3?

ReplyDeletewhat is the final answer?

ReplyDeleteatleast 1 refers to following possibilities for a 3 member team

ReplyDelete(1M + 2C) OR (2M + 1C) OR (3M)

6C1 X 6C2 + 6C1 X 6C2 + 6C3 = 200

No mehican, 6C3 = 20

ReplyDeleteAll = 12.11.10/6 = 220

220 - 20 = 200

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ReplyDeleteI concur with Bhushan's solution, but what is wrong with 6C1 X 11C2=330 (Means 1 Mexican for sure the other 2 can be picked from the remaining 11).

ReplyDeleteBlaoism's solution also makes sense:

12C3 - 6C3

Look at it in another way

M _ _ = 2nd spot can be filled 11 and 3rd 10 ways. That gives you 110.(Wouldnt this cover the 2M and 3M cases). I know the correct way is to branch it out as Bhushan and Blaoism have done.

Possible combination for the 3 delegates

ReplyDeleteM C C = M is 6C1 and C is 6C2 hence 6C1 + 6C2 = 90

M M C = C is 6C1 and M is 6C2 hence 6C1 + 6C2 = 90

C C C = C is 6C3 = 20

Add them together: 90+90+20 = 200