2
$\begingroup$

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!?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

We have to treat the two partitions say $(AB)(CD)(EFG)$ and $(CD)(AB)(EFG)$ as equal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.