different ways

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adi_800: Legendary Member; Posts: 544; Joined: Thu Oct 08, 2009 9:14 amFollowed by: 2 members; Location: Pune, India; Thanked: 31 times

by adi_800 » Fri Jul 30, 2010 1:53 am

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

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Source: Beat The GMAT — Problem Solving | Fri Jul 30, 2010 1:53 am

outreach: Legendary Member; Posts: 748; Joined: Sun Jan 31, 2010 7:54 amFollowed by: 3 members; Thanked: 46 times

by outreach » Fri Jul 30, 2010 2:12 am

A,B,C be the respective teams..
we have nine students, and we want to arrange them into teams of three.

so AAABBBCCC=9!/(3! 3! 3!) = 1680
but order of A,B,C can change hence 1680/3!=280

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kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members; Location: London; Thanked: 70 times

by kmittal82 » Fri Jul 30, 2010 2:29 am

I approached it differently, and no doubt got a wrong answer, but I would appreciate it if someone could explain my incorrect thinking here:

Out of 9 people, we can select any 3 in 9C3 = 84 ways

Out of the remaining 6 people, we can select 3 in 6C3 = 20 ways

Out of the remaining 3, we select 3 people in 3C3 = 1 way

Total ways of selecting = 84+20+1 = 105

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Fri Jul 30, 2010 2:31 am

Out of the nine people:
first three can be chosen in 9c3 ways = 84
second three can be chosen in 6c3 ways = 20
third three can be chosen in 3c3 = 1 ways

We want all 3 things to happen. Hence multiply = 84*20 = 1680.

Now, order of these groups doesnt matter. hence, 1680/3! gives the number of arrangements.
=280

hope this helps!!

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kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members; Location: London; Thanked: 70 times

by kmittal82 » Fri Jul 30, 2010 3:29 am

kvcpk wrote:Out of the nine people:
first three can be chosen in 9c3 ways = 84
second three can be chosen in 6c3 ways = 20
third three can be chosen in 3c3 = 1 ways

We want all 3 things to happen. Hence multiply = 84*20 = 1680.

Now, order of these groups doesnt matter. hence, 1680/3! gives the number of arrangements.
=280

hope this helps!!

ah, thanks for that

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ssd678: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Wed Aug 04, 2010 7:57 am

by ssd678 » Wed Aug 04, 2010 8:00 am

HI,

Could you explain why did you need to divide by 3!

kmittal82 wrote:
kvcpk wrote:Out of the nine people:
first three can be chosen in 9c3 ways = 84
second three can be chosen in 6c3 ways = 20
third three can be chosen in 3c3 = 1 ways

We want all 3 things to happen. Hence multiply = 84*20 = 1680.

Now, order of these groups doesnt matter. hence, 1680/3! gives the number of arrangements.
=280

hope this helps!!
ah, thanks for that

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Aug 07, 2010 2:07 am

adi_800 wrote: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

I was asked by PM to explain this problem.

When we're counting combinations -- when order doesn't matter -- we need a smaller answer choice than when we're counting arrangements.

Why? Because while ABC, ACB, BAC, BCA, CAB, CBA are all different arrangements of the letters ABC, they are the same combination. So the letters ABC give us 6 possible arrangements but only 1 possible combination. To account for all the duplicate combinations, we need to do some dividing:

total possible combinations = (total possible arrangements)/(number of elements being chosen)!

In the problem above, we have 9 people.

Group 1: (9*8*7)(3*2*1) = 84 possible groups. [9 choices for 1st position, 8 choices for 2nd, 7 choices for 3rd. The number of people being chosen is 3, so we divide by 3! to get rid of the duplicate combinations.]

Group 2: (6*5*4)/(3*2*1) = 20 possible groups. [6 choices for 1st position, 5 choices for 2nd, 4 choices for 3rd. The number of people being chosen is 3, so we divide by 3! to get rid of the duplicate combinations.]

Group 3: (3*2*1)/(3*2*1) = 1 possible group. [3 choices for 1st position, 2 choices for 2nd, 1 choice for 3rd. The number of people being chosen is 3, so we divide by 3! to get rid of the duplicate combinations.]

Now we need to combine our options for the 3 groups. There are two possible scenarios:

The order of the groupings matters:
If the order matters, we care whether ABC is assigned to group 1 or to group 3.
In this case, we just multiply the results above:
84*20*1 = 1680. [84 choices for group 1, 20 choices for group 2, 1 choice for group 3.]
This is the situation most likely to be discussed on the GMAT.

The order of the groupings doesn't matter:
If the order doesn't matter, we don't care whether ABC is assigned to group 1 or to group 3, because ABC - DEF - GHI gives us the same 3 groupings as DEF - GHI - ABC. In this case, we have to do some dividing to account for the duplicate groupings:
(84*20*1)/(3*2*1) = 280. [84 choices for Group 1, 20 choices for Group 2, 1 choice for Group 3. Since we don't care about the order of the 3 groups, we divide by 3! to get rid of the duplicate groupings.]
This scenario is unlikely to be discussed on the GMAT.

The problem above doesn't seem to care about the order of the groupings, so the correct answer is A.

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aarati: Senior | Next Rank: 100 Posts; Posts: 37; Joined: Thu Aug 05, 2010 3:15 am

by aarati » Mon Aug 09, 2010 12:20 am

GMATGuruNY wrote:
adi_800 wrote: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
I was asked by PM to explain this problem.

When we're counting combinations -- when order doesn't matter -- we need a smaller answer choice than when we're counting arrangements.

Why? Because while ABC, ACB, BAC, BCA, CAB, CBA are all different arrangements of the letters ABC, they are the same combination. So the letters ABC give us 6 possible arrangements but only 1 possible combination. To account for all the duplicate combinations, we need to do some dividing:

total possible combinations = (total possible arrangements)/(number of elements being chosen)!

In the problem above, we have 9 people.

Group 1: (9*8*7)(3*2*1) = 84 possible groups. [9 choices for 1st position, 8 choices for 2nd, 7 choices for 3rd. The number of people being chosen is 3, so we divide by 3! to get rid of the duplicate combinations.]

Group 2: (6*5*4)/(3*2*1) = 20 possible groups. [6 choices for 1st position, 5 choices for 2nd, 4 choices for 3rd. The number of people being chosen is 3, so we divide by 3! to get rid of the duplicate combinations.]

Group 3: (3*2*1)/(3*2*1) = 1 possible group. [3 choices for 1st position, 2 choices for 2nd, 1 choice for 3rd. The number of people being chosen is 3, so we divide by 3! to get rid of the duplicate combinations.]

Now we need to combine our options for the 3 groups. There are two possible scenarios:

The order of the groupings matters:
If the order matters, we care whether ABC is assigned to group 1 or to group 3.
In this case, we just multiply the results above:
84*20*1 = 1680. [84 choices for group 1, 20 choices for group 2, 1 choice for group 3.]
This is the situation most likely to be discussed on the GMAT.

The order of the groupings doesn't matter:
If the order doesn't matter, we don't care whether ABC is assigned to group 1 or to group 3, because ABC - DEF - GHI gives us the same 3 groupings as DEF - GHI - ABC. In this case, we have to do some dividing to account for the duplicate groupings:
(84*20*1)/(3*2*1) = 280. [84 choices for Group 1, 20 choices for Group 2, 1 choice for Group 3. Since we don't care about the order of the 3 groups, we divide by 3! to get rid of the duplicate groupings.]
This scenario is unlikely to be discussed on the GMAT.

The problem above doesn't seem to care about the order of the groupings, so the correct answer is A.

thanks,but is it one solution to solve these problem .......please explain me in different way to solve it..

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Aug 13, 2010 2:18 pm

aarati wrote:
GMATGuruNY wrote:
adi_800 wrote: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
thanks,but is it one solution to solve these problem .......please explain me in different way to solve it..
Here's another way to approach this problem:

The 1st person chosen can be put together with 8C2 = 28 pairs.
6 people left.
The next person chosen can be put together with 5C2 = 10 pairs.
3 people left.
The next person chosen can be put together with 2C2 = 1 pair.

28*10*1 = 280 total ways to group the people.

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