Hello, clockingly!

I think you mean "combinations".

The order of the names on a team is not considered.

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: ways.

There are 7 children; we choose 3 of them: ways.

Hence, there are: . ways to select team #1.

Select team #2.

There are 8 available adults; we choose 4 of them: ways.

There are 4 available children; we choose 2 of them: ways.

Hence, there are: . ways to select team #2.

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

Therefore, there are: . 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: .