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
different ways
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kmittal82
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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
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
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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!!
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
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ah, thanks for thatkvcpk 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!!
HI,
Could you explain why did you need to divide by 3!
Could you explain why did you need to divide by 3!
kmittal82 wrote:ah, thanks for thatkvcpk 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!!
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GMATGuruNY
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I was asked by PM to explain this problem.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
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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thanks,but is it one solution to solve these problem .......please explain me in different way to solve it..GMATGuruNY wrote:I was asked by PM to explain this problem.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
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 wrote:Here's another way to approach this problem:GMATGuruNY wrote:thanks,but is it one solution to solve these problem .......please explain me in different way to solve it..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
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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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
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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
