Wednesday, June 10, 2009

Problem 37 - GMAT Combinatorics

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

8 comments:

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

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  2. So you are saying 12 pick 3 minus 6 pick 3?

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  3. what is the final answer?

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  4. atleast 1 refers to following possibilities for a 3 member team

    (1M + 2C) OR (2M + 1C) OR (3M)
    6C1 X 6C2 + 6C1 X 6C2 + 6C3 = 200

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  5. No mehican, 6C3 = 20
    All = 12.11.10/6 = 220

    220 - 20 = 200

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  6. This comment has been removed by the author.

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  7. I 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).
    Blaoism'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.

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  8. Possible combination for the 3 delegates

    M 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

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