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

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?

Re: No. of ways to select 9 people

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

$\displaystyle 3!\binom{4}{2}\binom{4}{3}\binom{4}{4}+1!\binom{4} {3}\binom{4}{3}\binom{4}{3}=\ldots=208$

Re: No. of ways to select 9 people

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

You forgot a combination: three contestants from each group.

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

Too slow . . .