Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 05 Oct 2008
Posts: 273

In how many different ways can a group of 9 people be [#permalink]
Show Tags
29 Oct 2009, 05:30
8
This post was BOOKMARKED
Question Stats:
65% (01:44) correct
35% (01:07) wrong based on 293 sessions
HideShow timer Statistics
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: inhowmanydifferentwayscanagroupof9peoplebedivide101722.html
Official Answer and Stats are available only to registered users. Register/ Login.



Senior Manager
Joined: 18 Aug 2009
Posts: 320

Re: Probability [#permalink]
Show Tags
29 Oct 2009, 06:09
1
This post received KUDOS
2
This post was BOOKMARKED
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 = \(9!/(3!)^3*3! = 280\)



Math Expert
Joined: 02 Sep 2009
Posts: 39719

Re: Probability [#permalink]
Show Tags
29 Oct 2009, 06:09
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: \(\frac{9!}{(3!)^3*3!}=280\) You can also refer to the similar problem: combinationgroupsandthatstuff85707.html
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 18 Aug 2009
Posts: 299

Re: Probability [#permalink]
Show Tags
29 Oct 2009, 06:11
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) ?



Senior Manager
Joined: 18 Aug 2009
Posts: 299

Re: Probability [#permalink]
Show Tags
29 Oct 2009, 06:18
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)



Math Expert
Joined: 02 Sep 2009
Posts: 39719

Re: Probability [#permalink]
Show Tags
29 Oct 2009, 06:19



Intern
Joined: 09 Sep 2012
Posts: 33
Location: United States

Re: Probability [#permalink]
Show Tags
15 Oct 2012, 11:00
2
This post received KUDOS
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: \(\frac{9!}{(3!)^3*3!}=280\) You can also refer to the similar problem: combinationgroupsandthatstuff85707.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



Math Expert
Joined: 02 Sep 2009
Posts: 39719

Re: Probability [#permalink]
Show Tags
15 Oct 2012, 11:12
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: \(\frac{9!}{(3!)^3*3!}=280\) You can also refer to the similar problem: combinationgroupsandthatstuff85707.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: 6peopleformgroupsof2forapracticalworkeachgroup95344.htmlprobability85993.html?highlight=divide+groupscombination55369.html#p690842probability88685.html#p669025combinationgroupsandthatstuff85707.html#p642634subcommittee86346.html?highlight=divide+groupsHope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 09 Sep 2012
Posts: 33
Location: United States

Re: Probability [#permalink]
Show Tags
15 Oct 2012, 22:44
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: \(\frac{9!}{(3!)^3*3!}=280\) You can also refer to the similar problem: combinationgroupsandthatstuff85707.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: 6peopleformgroupsof2forapracticalworkeachgroup95344.htmlprobability85993.html?highlight=divide+groupscombination55369.html#p690842probability88685.html#p669025combinationgroupsandthatstuff85707.html#p642634subcommittee86346.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 3digit 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



Intern
Joined: 02 Nov 2012
Posts: 35

Re: In how many different ways can a group of 9 people be [#permalink]
Show Tags
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)



Intern
Joined: 03 May 2011
Posts: 10

Re: In how many different ways can a group of 9 people be [#permalink]
Show Tags
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)



CEO
Joined: 17 Nov 2007
Posts: 3584
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: In how many different ways can a group of 9 people be [#permalink]
Show Tags
30 May 2013, 15:43
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: mathcombinatorics87345.html
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



Math Expert
Joined: 02 Sep 2009
Posts: 39719

Re: In how many different ways can a group of 9 people be [#permalink]
Show Tags
31 May 2013, 01:56




Re: In how many different ways can a group of 9 people be
[#permalink]
31 May 2013, 01:56







