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good question..

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by Brent@GMATPrepNow » Wed Aug 21, 2013 11:26 am
rishianand7 wrote:In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

280
1260
1680
2520
3360
From the wording of the question, it's difficult to determine whether the 3 groups are distinct. This ambiguity makes the question unanswerable.

Are the 3 groups distinct? In other words, is being in group A different from being in group B?
If so, then the answer is C. However, if the 3 groups are not distinct, then the correct answer is A.

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by [email protected] » Wed Aug 21, 2013 12:23 pm
Hi rishianand7,

Brent is correct in that the question is poorly worded. I think that the "intent" of the question is this:

1st group = 9c3 = 9!/[3!6!] = 84 possibilities

2nd group = 6c3 = 6!/[3!3!] = 20 possibilities

3rd group = 3c3 = 1 possibility

Next, we have to account for the idea that any group could be first, second or third, so 123 is the same as 213, 312, etc. To remove the duplicates, we have to divide by 3!

So, (84x20x1)/3! = 280 groups

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by rishianand7 » Thu Aug 22, 2013 9:03 pm
Dear Rich and Brent,

The correct answer is in fact 280.
The question is a little ambiguous.

Thanks,
Rishi
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by GMATGuruNY » Thu Aug 22, 2013 9:35 pm
rishianand7 wrote:In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

280
1260
1680
2520
3360
An alternate approach:

The FIRST PERSON selected must be combined with a PAIR of people to form a group of 3.
From the 8 OTHER PEOPLE, the number of pairs that be formed = 8C2 = (8*7)/(2*1) = 28.

6 people left.
The NEXT PERSON selected must also be combined with a PAIR of people to form a group of 3.
From the 5 OTHER PEOPLE, the number of pairs that can be formed = 5C2 = 10.

3 people left.
The NEXT PERSON selected must also be combined with a PAIR of people to form a group of 3.
From the 2 OTHER PEOPLE, the number of pairs that can be formed = 2C2 = 1.

To combine these options, we multiply:
28*10*1 = 280.
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