Anagram Method for Combinations - Does it break here?

[joel.vidalphillips]

I came across a problem where I tried to apply the anagram method and it didn't work. I'm wondering if it's because of the 3 equal groups part of the question...


Question:
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

a) 280
b) 1260
c) 1680
d) 2520
e) 3360

If you choose 3 different letters to represent the 3 groups (A, B, C) and
If you set up 9 different slots: 9 8 7 6 5 4 3 2 1
A A A B B B C C C

it seems like you should be able to solve this problem by saying 9!/(3! * 3! * 3!)

This however does not result in the correct answer which is (a).

After reading the answer explanation i get it, but I just don't get why the anagram method breaks down here. If someone could provide an explanation so I can better use the method in the future (or avoid using it if there are exceptions to when it can be used) that would be great.

Thanks.

-Joel

[mikhsor]

It actually works. When you get 1680 you just have to divide it by 3! to account for different orders of choosing 3 groups out of 9 people. 1680/2/3=280.

Let's say we are choosing 3 groups: A, B, and C. It doesn't matter if we choose A first or B first, right? that is why we need to divide by 3!

[joel.vidalphillips]

Yeah, I understand you have to divide by 3! again, but it kinda breaks the anagram analogy. You have to take an additional step to get the anagram calculation to work.

I just wanted to see if there were a better explanation or an exception to the rule and not just "it works, just divide it again by 3!"

Appreciate the answer though.

[mikhsor]

This problem is a little more complicated than just simple anagram method because you are choosing 3 groups of elements each containing 3 elements and the order is not important. Look at it this way. If you had 3 elements to choose from (A,B,C) in how many ways could you arrange them? ABC, ACB, BAC, BCA, CAB, CBA - 6 times. This problem, however, states that each one of these elements (A,B,C) should contain 3 other elements, which you also can arrange in different ways within each group. Or even better:
we have 9 elements A,B,C,D,E,F,G,H,I. Let's say we want to have ABC in the one group, DEF in another, and GHI in the last one. So, does it matter to you how you choose elements? No. Let's see:

ABC DEF GHI or DEF GHI ABC or DEF GHI ABC, etc. are all the same, the order doesn't matter... Hope it is helpful...

[mikhsor]

Got even better idea how to help you understand :-)

see, you are saying:
"If you choose 3 different letters to represent the 3 groups (A, B, C) and
If you set up 9 different slots: 9 8 7 6 5 4 3 2 1
A A A B B B C C C " and then you give your solution (9!/3!*3!*3!). According to this approach the following sequences are different:

A B A C B A B C C and C B A A B A B C C . They would be different if, for example, they were two words. In this problem, however, these sequces consist of 3 separate clusters. To make the second "word" I just moved the first cluster (ABA) to the place between CBA and BCC. In the context of this problem these two sequences are the same because I just mixed the groups (if they were two words, they of course would be different and you would have to count each of them). And there are 3! ways you can mix these clusters within 9 letter "word"

[Ben Ku]

Joel,

As you and the other responders have figured out, this happens to be a bit more complicated than a typical combinations problem.

Your first step is good. Let's suppose we have nine students, and we want to arrange them into teams of three. It would basically be an anagram of AAABBBCCC, or 9!/(3! 3! 3!) = 1680.

However, it doesn't matter that the first team is team A, the second team is team B, and the third team is team C. If we had an arrangement AAABBBCCC, this other arrangement CCCAAABBB is exactly the same because the same people are in the same teams. Because of this, we have to divide 1680 by 3! because there are 3! ways to arrange ABC. The answer should be 280.

I hope that makes sense.