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Math Help - Need help with a few questions.

  1. #16
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    Re: Need help with a few questions.

    So we choose the 3 suits our hand consists of: 4C3.

    Now we choose 5 of the 13 cards of the first suit: 13C5

    Now we choose 6 of the 13 cards of the second suit: How do we do this?

    Now we choose 2 of the 13 cards of the last suit: How do we do this?
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  2. #17
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    Re: Need help with a few questions.

    Quote Originally Posted by Jame View Post
    So we choose the 3 suits our hand consists of: 4C3.

    Now we choose 5 of the 13 cards of the first suit: 13C5

    Now we choose 6 of the 13 cards of the second suit: How do we do this?

    Now we choose 2 of the 13 cards of the last suit: How do we do this?
    So 1287, 1716, 78.

    Then multiply them all to get a hand?
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  3. #18
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    Re: Need help with a few questions.

    Yes, we multiply

    4C3 * 13C5 * 13C6 * 13C2
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  4. #19
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    Re: Need help with a few questions.

    Quote Originally Posted by Jame View Post
    Yes, we multiply

    4C3 * 13C5 * 13C6 * 13C2
    Ok I got 689 049 504, which is the final answer?
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  5. #20
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    Re: Need help with a few questions.

    Yep, that is the answer. You can also leave it 4C3 * 13C5 * 13C6 * 13C2 if you want.
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  6. #21
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    Re: Need help with a few questions.

    Quote Originally Posted by Jame View Post
    Yep, that is the answer. You can also leave it 4C3 * 13C5 * 13C6 * 13C2 if you want.
    Okay, thanks so much for the help, both you and emakarov!
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  7. #22
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    Re: Need help with a few questions.

    I'm sorry I was actually mistaken in my answer for 2). Since we have different amounts of each suit (5,6,2), we must choose each suit individually. So instead of 4C3, its 4*3*2 (or 4!)

    The rest of the answer is the same.

    Example: a hand with "5 hearts, 6 spades and 2 diamonds" is different then a hand with "5 diamonds, 6 hearts and 2 spades"
    (order of the suits does in fact matter here)

    If we had a different breakdown of suits (like 5,5,3). We would have to choose the two suits with the 5 cards (4C2), then choose the suit with 3 cards (2C1)

    The only time we would choose the suits all together is if each card had the same breakdown (e.g. (3,3,3))
    Last edited by Jame; February 24th 2013 at 08:08 PM.
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  8. #23
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    Re: Need help with a few questions.

    Quote Originally Posted by Jame View Post
    I'm sorry I was actually mistaken in my answer for 2). Since we have different amounts of each suit (5,6,2), we must choose each suit individually. So instead of 4C3, its 4*3*2 (or 4!)

    The rest of the answer is the same.

    Example: a hand with "5 hearts, 6 spades and 2 diamonds" is different then a hand with "5 diamonds, 6 hearts and 2 spades"
    (order of the suits does in fact matter here)

    If we had a different breakdown of suits (like 5,5,3). We would have to choose the two suits with the 5 cards (4C2), then choose the suit with 3 cards (2C1)

    The only time we would choose the suits all together is if each card had the same breakdown (e.g. (3,3,3))
    Changed, thanks for letting me know!
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