dividing into equal groups (1 Viewer)

[laters]

"How many ways can 21 people be divided into 3 equal groups?"

Just wondering if the answer should be 21C7 14C7 7C7 or same but divided by 3!, seeing as I don't know if the groups are implied to be 'distinct' or not.

 

[Shadowdude]

1. Choose 7 people to go into the first group: C(21,7) ways
2. Choose 7 people to go into the second group from the remaining 14: C(14,7) ways
3. The last 7 go into the last group: 1 way = C(7,7)
4. Divide by 3! as the groups are not distinct/unique


So yes, you need to divide by 3!

 

[Drongoski]

Isn't this just = 21!/(7! x 7! x 7!) - the number of ways of partitioning 21 distinct objects into 3 groups of 7 each ???

 

[InteGrand]

Isn't this just = 21!/(7! x 7! x 7!) - the number of ways of partitioning 21 distinct objects into 3 groups of 7 each ???

This is indeed equivalent to laters' expression of (21C7)*(14C7), as we can see by writing each of these in factorial form and cancelling stuff from numerators and denominators of adjacent fractions.

In terms of laters' main Q. (which was whether the groups are 'labelled' or not), I'd guess we should treat them as NON-labelled unless otherwise stated. You could write that you have assumed this so the marker knows you did the question right according to your interpretation of it. If you wanted you could also say that if the groups are meant to be labelled, like Groups A, B and C say, we wouldn't divide by the 3!.