Combinations: form 3 teams from 30 people

Question

In how many ways can you form 3 teams from 30 people where each team consists of 1 team leader and 9 other members?

Attempted solution

I need to pick 3 team leaders and the order is unimportant and there are no repetitions:

$C_{30}^3$=4060

For the first team I have 27 people to choose from and I need to pick 9. Order is not important and there are no repetitions:

$C_{27}^9$=4686825

For the second team I have 18 people left and I need to pick another 9:

$C_{18}^9$=48620

For the third team I have 9 people left:

$C_{9}^9$=1

To get the solution, I just multiply all results:

$4060*4686825*48620*1=925166131890000$

Is it correct?

Answer 1

There are $\dbinom{30}{3}$ ways of selecting the team leaders. Now line up the team leaders in order of height.

The tallest team leader can select her team in $\dbinom{27}{9}$ ways. For each of these ways, the next tallest team leader can select her team in $\dbinom{18}{9}$ ways. Multiply.

Another way: We will first select the teams. For each selection, the team leaders can be then chosen in $10^3$ ways.

To find the number of ways of selecting the teams, let us first change the problem to selecting three teams, to wear respectively blue, white, and red uniforms. The blue uniform people can be chosen in $\dbinom{30}{10}$ ways. For each of these ways, the white uniform people can be chosen in $\dbinom{20}{10}$ ways, and we are finished.

But for each way to select the teams, there are $3!$ ways to assign the uniforms. Thus the total number of ways to select naked teams with leaders is $$\frac{1}{3!}\binom{30}{10}\binom{20}{10}10^3.$$

Answer 2

If you have three teams at the beginning, and assign the 30 people into the groups:

First, choose the members of the three teams. So there should be $C_{30}^{10}\times C_{20}^{10}$ ways of doing that. Then, choose a leader for each team and there are $10^3$ ways of doing this.

However, if you just want to separate the 30 people into 3 groups, your approach is correct, which is $\frac{1}{6}$ of the other case.