# Thread: No. of ways to select 9 people

1. ## 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

2. ## 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?

3. ## 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:

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

4. ## 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: $\displaystyle (_4C_4 \times _4C_3 \times _4C_2) \times 6\:=\:144$, 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 . . .