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Math Help - Possibilities? (I don't remember the exact word)

  1. #1
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    Possibilities? (I don't remember the exact word)

    Hi! First time poster.

    So I'm doing some homework, due in six hours. I'm almost done, but there's a couple of questions I'm not too sure of: one has to do with possibilities, the other not so much.

    1. The easier one:
    What is the coefficient of X^11 in (3 + X)^15?

    I'm guessing it's 15!/11!4! = 1365 but that 3 in (X + 3)is making me uncomfortable.

    The one that's bothering me:
    2. Suppose that a class contains 12 men and 17 women.
    (a) How many different 10 person teams can be made?
    (b) How many different 5 person teams can be made that contain at least two men?

    (a) Is it 19^10? That's the only thing I can think of (19 = 12 + 17) but for some reason I don't feel right about it.
    (b) "At least". That "at least" changes everything. Any help would be appreciated.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Dorcon View Post
    Hi! First time poster.

    So I'm doing some homework, due in six hours. I'm almost done, but there's a couple of questions I'm not too sure of: one has to do with possibilities, the other not so much.

    1. The easier one:
    What is the coefficient of X^11 in (3 + X)^15?

    I'm guessing it's 15!/11!4! = 1365 but that 3 in (X + 3)is making me uncomfortable.

    The one that's bothering me:
    2. Suppose that a class contains 12 men and 17 women.
    (a) How many different 10 person teams can be made?
    (b) How many different 5 person teams can be made that contain at least two men?

    (a) Is it 19^10? That's the only thing I can think of (19 = 12 + 17 Mr F butts in: You're kidding, right?) but for some reason I don't feel right about it.
    (b) "At least". That "at least" changes everything. Any help would be appreciated.

    Thanks in advance!
    1. The general term is {15 \choose r} 3^{15 - r} x^r. You require the coefficient of x^{11} \, ....

    2. (a) {29 \choose 10}.

    (b)  {12 \choose 2} \cdot {17 \choose 3} + {12 \choose 3} \cdot {17 \choose 2} + {12 \choose 4} \cdot {17 \choose 1} + {12 \choose 5} \cdot {17 \choose 0}.
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  3. #3
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    Smile

    OK, excellent. Thanks.

    I don't suppose the answer to the first question would be 110565 would it?
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  4. #4
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    Binomial Coefficients

    Hello Dorcon
    Quote Originally Posted by Dorcon View Post
    OK, excellent. Thanks.

    I don't suppose the answer to the first question would be 110565 would it?
    It certainly would!

    Grandad
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