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In how many different ways can a group of 9 people be [#permalink]
 29 Oct 2009, 05:30
Question Stats:
56% (02:07) correct
43% (00:56) wrong based on 7 sessions
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
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Bunuel wrote: study wrote: is there a simple way to solve this:
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 our case mn=9, m=3 groups and n=3 people: \frac{9!}{(3!)^3*3!}=280You can also refer to the similar problem: combination-groups-and-that-stuff-85707.htmlHi Bunuel, why are we dividing it by 3! ? when the order matters we multiply 9C3*6C3*3C3 by 3! right ? please clear my concept for this ? Thanks
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study wrote: is there a simple way to solve this:
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 our case mn=9, m=3 groups and n=3 people: \frac{9!}{(3!)^3*3!}=280You can also refer to the similar problem: combination-groups-and-that-stuff-85707.html
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Number of ways 9 people can be arranged = 9!
Number of ways 3 people within a group can be arranged = 3!
Number of ways 3 groups can be arranged = 3!
Number of ways 9 people can be divided in group of 3 people = 9!/(3!)^3*3! = 280
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My attempt:
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)
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gmattokyo wrote: My attempt:
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)
? I stand corrected  as order doesn't matter 1680/3!=280 (A)
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154238 wrote: Bunuel wrote: study wrote: is there a simple way to solve this:
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 our case mn=9, m=3 groups and n=3 people: \frac{9!}{(3!)^3*3!}=280You can also refer to the similar problem: combination-groups-and-that-stuff-85707.htmlHi Bunuel, why are we dividing it by 3! ? when the order matters we multiply 9C3*6C3*3C3 by 3! right ? please clear my concept for this ? Thanks No, in this case we just don't divided by 3!. Check these links for more: 6-people-form-groups-of-2-for-a-practical-work-each-group-95344.htmlprobability-85993.html?highlight=divide+groupscombination-55369.html#p690842probability-88685.html#p669025combination-groups-and-that-stuff-85707.html#p642634sub-committee-86346.html?highlight=divide+groupsHope it's clear.
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
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154238 wrote: Bunuel wrote: study wrote: is there a simple way to solve this:
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 our case mn=9, m=3 groups and n=3 people: \frac{9!}{(3!)^3*3!}=280You can also refer to the similar problem: combination-groups-and-that-stuff-85707.htmlHi Bunuel, why are we dividing it by 3! ? when the order matters we multiply 9C3*6C3*3C3 by 3! right ? please clear my concept for this ? Thanks No, in this case we just don't divided by 3!. Check these links for more: 6-people-form-groups-of-2-for-a-practical-work-each-group-95344.htmlprobability-85993.html?highlight=divide+groupscombination-55369.html#p690842probability-88685.html#p669025combination-groups-and-that-stuff-85707.html#p642634sub-committee-86346.html?highlight=divide+groupsHope it's clear. 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! !!! Please correct if i am wrong ? Thanks
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Re: In how many different ways can a group of 9 people be [#permalink]
04 Jan 2013, 07:27
Can someone write it down in faculty forms instead of combinatorics (because I do not know how to use them on the GMAT without a calculator)
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Re: In how many different ways can a group of 9 people be
[#permalink]
04 Jan 2013, 07:27
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