Combinations and Committees

[lacrotix]

Hi there. I have a question about combinations at the Math 12 Principles level.

Homework Statement

From a group of 5 student representatives, 3 will be chosen to work on the dance committee. Also, each committee must have a chairperson. How many committees are possible?

The Attempt at a Solution

So I solved this question with
5 nCr 3 x 3 nCr 1 = 30
In which the first part of the equation finds all the possible committees that can be made, and the second part of the equation lists how many different leaders can be chosen from the sub-committee of three I just made.However, I don't see why I can't use this:
5 nCr 2 x 5 nCr 1 = 50
Where the first part chooses a combination of two followers and the second part chooses a combination of one leader for the sub-committee of three people.

Any ideas?

 

[HallsofIvy]

Instead of trying to use a formula think of it this way: There are, initially, 5 people to choose the first person from so there are 5 ways to do that. After the first person is chosen there are 4 people left to choose from so 4 ways to choose the second person. Finally, there are 3 people left to choose from. There would be 5(4)(3)= 60 ways to do that. However, we can count the first person as the chairman so the order in which the last two are chosen is not important. There are 2!= 2 ways that the same 2 people could has been chosen different orders so divide by 2.

As for the formulas, that is the same as
$$\frac{5(4)(3)}{2!}= \frac{5(4)(3)(2)(1)}{2!2!}= \frac{5!}{2!2!}$$
which is the same as
$$\frac{5!}{2!3!}\frac{3!}{1!2!}= _5C_3 _3C_2$$
as you say.

Your error is that you are choosing two people out of 5 to be "members" and then one of the same 5 people to be chair. That makes it possible that one of the two chosen before will also be chosen chair, leaving the committee with only two members. Instead, you want to chose 2 "members" from the 5 people, then one chair from the remaining three people who were not already chosen. That gives
$$_5C_2 _3C_1= \frac{5!}{3!2!}\frac{3!}{2!1!}= \frac{5!}{2!2!}= _5C_3 _3C_2$$
as before.

 

[lacrotix]

Thanks for the quick response HallsofIvy.

However, my question relates to the last paragraph of my question. I do understand how to get it your way, but I don't understand how my original thinking is wrong.

 

[HallsofIvy]

Yes, while you were reading I was editing my answer. You say

However, I don't see why I can't use this:
5 nCr 2 x 5 nCr 1 = 50
Where the first part chooses a combination of two followers and the second part chooses a combination of one leader for the sub-committee of three people.

"one leader for the subcommittee of three people" would be $_3C_1$
not $_5C_1$

 

[lacrotix]

Thanks. That does help a whole lot!