# combinations problem

• Dec 15th 2006, 06:51 PM
clockingly
combinations problem
ok, so i'm having trouble with the following problem, so if anyone could help, i'd be grateful. i think that there's some kind of permutation needed for this (meaning n choose k), but i'm confused about how to use it.

A bowling team has 11 adults and 7 children. One night, they decide to divide themselves into 3 teams. The first will have 3 adults and 3 children. The second will have 4 adults and 2 children. The third will also have 4 adults and 2 children. In how many ways can the three teams be chosen?
• Dec 16th 2006, 03:57 AM
Soroban
Hello, clockingly!

I think you mean "combinations".
The order of the names on a team is not considered.

Quote:

A bowling team has 11 adults and 7 children.
One night, they decide to divide themselves into 3 teams.
. . The first will have 3 adults and 3 children.
. . The second will have 4 adults and 2 children.
. . The third will also have 4 adults and 2 children.

In how many ways can the three teams be chosen?

I will assume that the three teams are ordered and distinct.
That is, that they are called "Team #1", "Team #2", and "Team #3". **

Select team #1.
There are 11 adults; we choose 3 of them: $\displaystyle \binom{11}{3}$ ways.
There are 7 children; we choose 3 of them: $\displaystyle \binom{7}{3}$ ways.
Hence, there are: .$\displaystyle \binom{11}{3}\binom{7}{3}\:=\:165\cdot35 \:=\:5,775$ ways to select team #1.

Select team #2.
There are 8 available adults; we choose 4 of them: $\displaystyle \binom{8}{4}$ ways.
There are 4 available children; we choose 2 of them: $\displaystyle \binom{4}{2}$ ways.
Hence, there are: .$\displaystyle \binom{8}{4}\binom{4}{2} \:=\:70\cdot6\:=\:420$ ways to select team #2.

The remaining 4 adults and 2 children will belong to team #3.

Therefore, there are: .$\displaystyle 5,775 \times 420 \:=\:\boxed{2,425,500}$ ways to select the teams.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** If teams #2 and #3 are interchangeable,
. . . then the "420" is twice as large as it should be.

The number of ways would be: .$\displaystyle 5,775 \times 210 \:=\:\boxed{1,212,750}$

• Dec 16th 2006, 06:53 AM
clockingly
Thank you so much! You explained that in a very clear way.