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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
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?
Another way: we can choose the nine contestants according to the outlineor
(up to permutations of the addends), so the solution is:
Hello, Punch!
Quote:
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:, but the answer is 208.
You forgot a combination: three contestants from each group.
. .more selections.
Too slow . . .