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  1. #1
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    combination

     8 bottles of zinfandel

     10 bottles of merlot

     12 bottles of cabaret.

    (a) If he wants to serve  3 bottles of zinfandel and serving order is important, how many ways are there to do this?

     \frac{8!}{5!} = 336


    (b) If  6 bottles of wine are to be randomly selected from the  30 , how many ways are there to do this?

     \binom{30}{6} = 593,775

    (c) If  6 bottles are randomly chosen, how many ways are there two bottles of each variety being chosen?


    (d) If  6 bottles are randomly chosen, what is the probability that this results in two bottles of each variety being chosen.

    (e) If  6 bottles are randomly chosen, what is the probability that they are all the same variety?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shilz222 View Post

    (c) If  6 bottles are randomly chosen, how many ways are there two bottles of each variety being chosen?
    { 8 \choose 2} {10 \choose 2} {12 \choose 2}

    (d) If  6 bottles are randomly chosen, what is the probability that this results in two bottles of each variety being chosen.
    \frac {{ 8 \choose 2} {10 \choose 2} {12 \choose 2}}{{ 30 \choose 6}}

    (e) If  6 bottles are randomly chosen, what is the probability that they are all the same variety?
    \frac {{ 8 \choose 6} +  {10 \choose 6} + {12 \choose 6}}{{30 \choose 6}}

    why is that?
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  3. #3
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    There are  \binom{8}{6} ways to choose all zinfandel, etc..
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