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Math Help - No. of ways to select 9 people

  1. #1
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    No. of ways to select 9 people

    12 contestants participate in a TV show. They are distributed into the morning, afternoon and evening sessions with 4 contestants in each session. During the show, nine contestants are selected to enter the final round, with at least 2 contestants from each of the three sessions. Find the number of ways this can be done.

    My answer is: (4C4 x 4C3 x 4C2) x 6=144 but answer is 208
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  2. #2
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    Re: No. of ways to select 9 people

    1) 12 choose 4 to separate off the first session
    2) 8 choose 4 to split the second and third sessions
    3) 4 choose 2 to get the first two contestants from the first session
    4) 4 choose 2 " " " second session
    5) 4 choose 2 " " " third session
    6) out of 6 remaining peeps, we need 3 from any, so 6 choose 3 for the last contestants

    All of these things must happen at the same time so multiply. See how I approached that?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: No. of ways to select 9 people

    Another way: we can choose the nine contestants according to the outline 9=2+3+4 or 9=3+3+3 (up to permutations of the addends), so the solution is:

    3!\binom{4}{2}\binom{4}{3}\binom{4}{4}+1!\binom{4}  {3}\binom{4}{3}\binom{4}{3}=\ldots=208
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  4. #4
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    Re: No. of ways to select 9 people

    Hello, Punch!

    12 contestants participate in a TV show.
    They are distributed into the morning, afternoon and evening sessions
    with 4 contestants in each session.
    During the show, nine contestants are selected to enter the final round,
    with at least 2 contestants from each of the three sessions.
    Find the number of ways this can be done.

    My answer is: (_4C_4 \times  _4C_3 \times  _4C_2) \times 6\:=\:144, but the answer is 208.

    You forgot a combination: three contestants from each group.

    . . _4C_3 \times _4C_3 \times _4C_3 \;=\;64 more selections.


    Too slow . . .
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