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

280 1,260 1,680 2,520 3,360

\frac{9C3*6C3*3C3}{3!}=280

(Dividing by 3! as the order doesn't matter.)

Actually there is a formula for it a bit complicated, but still if you need it, here you go:

The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \frac{(mn)!}{(n!)^m*m!}.

Total ways to form 3 groups= #of ways to form 1st group x #of ways to form 2nd group x 1 (3 people left after two groups are formed)

= 9C3 x 6C3 x 1 = 84 x 20 = 1680

(C) ?

Yes, but you don't have the first, second or second group. You have just groups of three. So, shouldn't you divide this by 3!? 1680/3!=280. _________________

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

280 1,260 1,680 2,520 3,360

\frac{9C3*6C3*3C3}{3!}=280

(Dividing by 3! as the order doesn't matter.)

Actually there is a formula for it a bit complicated, but still if you need it, here you go:

The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \frac{(mn)!}{(n!)^m*m!}.

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

280 1,260 1,680 2,520 3,360

\frac{9C3*6C3*3C3}{3!}=280

(Dividing by 3! as the order doesn't matter.)

Actually there is a formula for it a bit complicated, but still if you need it, here you go:

The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \frac{(mn)!}{(n!)^m*m!}.

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

280 1,260 1,680 2,520 3,360

\frac{9C3*6C3*3C3}{3!}=280

(Dividing by 3! as the order doesn't matter.)

Actually there is a formula for it a bit complicated, but still if you need it, here you go:

The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \frac{(mn)!}{(n!)^m*m!}.

It still not clear to me. Can you please clear that how the order is included in 9C3*6C3*3C3 ? Usually what we do is let say we have to make 3-digit numbers with the digits 1,4,7,8 and 9 if the digits are not repeated. ==>>

so 5C3 *3! (as order matters) = 60

so similar way if we are making 3 subsets and order matters then we have to multiply by 3! !! And if order doesn't matter then we should not divide by 3! !!!

Re: In how many different ways can a group of 9 people be [#permalink]
30 May 2013, 13:36

I too am not understanding how we are accounting for the order.

I got the numerator correct (9c3x6c3x3c3), but I don't understand why we divide it by 3!? I know we do it for the arrangements, but how are we accounting for the order in the first place?

I understand that if you have group 1 (g1), g2, and g3 that it is not asking for the arrangement of the groups and therefore order is not important. I still do not see why I would divide by 3!...

If order did matter then i would multiply by 3! to account for the arrangement, right?

My only other thought on this is that perhaps, since we account for an arrangement by 3! that 9c3, 6c3, and 3c3 is equivalent to 3 groups x 2 groups x 1 group since we are already multiplying, thus we account for this by dividing by 2 since there is ultimately a basic order that the 3 will have to create. The final division by 3 eludes me...

Or is it that the fact that we have multiplied all 3 scenarios means we have taken arrangement into account and need to take it back out creating the need to divide by 3!?

*sigh*...maybe I just stick with the other formula (mn/((n^m)*m)