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
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good question..
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rishianand7
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Brent@GMATPrepNow
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From the wording of the question, it's difficult to determine whether the 3 groups are distinct. This ambiguity makes the question unanswerable.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
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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Brent
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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
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rishianand7
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Dear Rich and Brent,
The correct answer is in fact 280.
The question is a little ambiguous.
Thanks,
Rishi
The correct answer is in fact 280.
The question is a little ambiguous.
Thanks,
Rishi
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An alternate approach: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
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
