How many ways can you divide 7 students into three teams where two of the teams contain two people and one of the teams contains three?
I got $\frac{7!}{2!2!3!}$ but the correct answer was $\frac{7!}{2!3!2!2!}$.
Why is there an extra 2!?
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You thought right just a bit incomplete. The answer you got shows that that you divide seven students in
(7!)/(2!2!3!).
Suppose the 2 groups which have 2 people each is labeled as G1 and G2.
Now the labeling could have been G2 and G1 also.
There are 2! ways in which you can label the groups having 2 people each !
If there had been say 3 groups of a certain number of students where each group had 3 people. Then in the denominator you would have got
3!3!3!3! .
The extra 3! comes from the 6 possible arrangements of the 3 groups themselves.
Like say you have ABCDEF and you have to distribute this into 3 groups of 2 each.
One such group be
AB CD EF
Now this group can be also written as
CD EF AB , EF AB CD , CD AB EF , AB EF CD , EF CD AB. ( 3! =6 Ways )
Similarly for each possible combination of groups.
That's why you get an extra 2! ( or 3! ) term due to arrangement of the groups.
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We have to treat the two partitions say $(AB)(CD)(EFG)$ and $(CD)(AB)(EFG)$ as equal.
