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In how many different ways can a group of 9 people be [#permalink]
 29 Oct 2009, 05:30
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41% (02:18) correct
58% (01:06) wrong based on 95 sessions
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people? A. 280 B. 1,260 C. 1,680 D. 2,520 E. 3,360 OPEN DISCUSSION OF THIS QUESTION IS HERE: in-how-many-different-ways-can-a-group-of-9-people-be-divide-101722.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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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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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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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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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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!! ,11 Mixed Questions NEW!!!, 12 Fresh Meat NEW!!!
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!!!, 11 New DS 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]
30 May 2013, 14:36
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)
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Re: In how many different ways can a group of 9 people be [#permalink]
30 May 2013, 15:43
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For those who don't understand why we need to divide by 3! Let say we have 3 people in blue pants, 3 in red pants, and 3 in green pants. C^9_3*C^6_3*C^3_3 have all following 6 combinations: [P P P] [P P P] [P P P][P P P] [P P P] [P P P][P P P] [P P P] [P P P][P P P] [P P P] [P P P][P P P] [P P P] [P P P][P P P] [P P P] [P P P]But in fact all those combinations represent only 1 outcome and that is why we need to exclude order (divide by 3!). Here is a quick review of fundamentals: math-combinatorics-87345.html
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Re: In how many different ways can a group of 9 people be [#permalink]
31 May 2013, 01:56
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Re: In how many different ways can a group of 9 people be
[#permalink]
31 May 2013, 01:56
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