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shankar.ashwin
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GMATGuruNY
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Number of combinations of 3 that can be formed from 7 students = 7C3 = 35.mariah wrote:In how many ways are possible to divide 7 students in 3 groups. if two groups got 2 people and another one 3 !
Number of ways to DIVIDE the remaining 4 students into pairs = (4C2 * 2C2)/2! = 3.
To combine the options above, we multiply:
35*3 = 105.
Please note the following distinction:
In how many ways can 4 students be ASSIGNED to 2 projects if each project must be ASSIGNED a pair of students and no student can work on more than 1 project?
4C2 * 2C2 = 6.
Here is what we have just counted:
Project 1 - Project 2:
AB - CD
AC - BD
AD - BC
CD - AB
BD - AC
BC - AD
Number of options = 6.
In how many ways can 4 students be DIVIDED into pairs?
(4C2 * 2C2)/2! = 3.
Here is what we have just counted:
AB-CD
AC-BD
AD-BC
Number of options = 3.
Notice that the pairings in bold red above are not included in our tally.
The number of unique pairings is smaller because AB-CD and CD-AB are 2 ways to ASSIGN the pairs but only ONE way to DIVIDE the 4 people into pairs.
The ORDER of the pairings DOESN'T MATTER.
To remove the duplicate pairings from our tally -- since ORDER DOESN'T MATTER -- we divide by the number of ways to ARRANGE the 2 pairs (2!).
For another problem about DIVIDING people into groups, check here:
https://www.beatthegmat.com/forming-teams-t73034.html
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GmatguruNY, Hi
A professor assigns three projects to seven students. The students will be divided two three groups, which has three, two, two students, respectively. How many ways are possible?
as you nicely explained answer for this question would be 210, - 7c3*4c2*2c2=210
but , I can't get question below, why answer is 620? Why we have to multiply 210 by 3, but not in first question ?
A professor will assign seven projects to three students. If two students each got 2 projects, and the other one got 3 projects, how many ways are possible?
A professor assigns three projects to seven students. The students will be divided two three groups, which has three, two, two students, respectively. How many ways are possible?
as you nicely explained answer for this question would be 210, - 7c3*4c2*2c2=210
but , I can't get question below, why answer is 620? Why we have to multiply 210 by 3, but not in first question ?
A professor will assign seven projects to three students. If two students each got 2 projects, and the other one got 3 projects, how many ways are possible?
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shankar.ashwin
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I think the answer should be 630.
210 as got before, and here there are 3 cases where any one of the students could get 3 projects, So 210*3 = 630.
Not sure why it would be 620
210 as got before, and here there are 3 cases where any one of the students could get 3 projects, So 210*3 = 630.
Not sure why it would be 620
mariah wrote:GmatguruNY, Hi
A professor assigns three projects to seven students. The students will be divided two three groups, which has three, two, two students, respectively. How many ways are possible?
as you nicely explained answer for this question would be 210, - 7c3*4c2*2c2=210
but , I can't get question below, why answer is 620? Why we have to multiply 210 by 3, but not in first question ?
A professor will assign seven projects to three students. If two students each got 2 projects, and the other one got 3 projects, how many ways are possible?
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InguteInga
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GMATGuruNY wrote:Number of combinations of 3 that can be formed from 7 students = 7C3 = 35.mariah wrote:In how many ways are possible to divide 7 students in 3 groups. if two groups got 2 people and another one 3 !
Number of ways to DIVIDE the remaining 4 students into pairs = (4C2 * 2C2)/2! = 3.
To combine the options above, we multiply:
35*3 = 105.
Please note the following distinction:
In how many ways can 4 students be ASSIGNED to 2 projects if each project must be ASSIGNED a pair of students and no student can work on more than 1 project?
4C2 * 2C2 = 6.
Here is what we have just counted:
Project 1 - Project 2:
AB - CD
AC - BD
AD - BC
CD - AB
BD - AC
BC - AD
Number of options = 6.
In how many ways can 4 students be DIVIDED into pairs?
(4C2 * 2C2)/2! = 3.
Here is what we have just counted:
AB-CD
AC-BD
AD-BC
Number of options = 3.
Notice that the pairings in bold red above are not included in our tally.
The number of unique pairings is smaller because AB-CD and CD-AB are 2 ways to ASSIGN the pairs but only ONE way to DIVIDE the 4 people into pairs.
The ORDER of the pairings DOESN'T MATTER.
To remove the duplicate pairings from our tally -- since ORDER DOESN'T MATTER -- we divide by the number of ways to ARRANGE the 2 pairs (2!).
For another problem about DIVIDING people into groups, check here:
https://www.beatthegmat.com/forming-teams-t73034.html
@GMATGuruNY
Looking at your explanation it seems that the method you used in this post here (https://www.beatthegmat.com/forming-teams-t73034.html) cannot be applied to this problem. Is that right? I am getting slightly confused. I though we can use the same principal: 1st person can be teamed up with another 6 and then another 5, so 6*5 to choose a team of 3. And then 4th person can be teamed up with another 3, while 6th person - with only 1. In total: 6*5*3...
I would much appreciate your help! Thank you!!
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GMATGuruNY
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https://www.beatthegmat.com/forming-teams-t73034.htmlInguteInga wrote: @GMATGuruNY
Looking at your explanation it seems that the method you used in this post here (https://www.beatthegmat.com/forming-teams-t73034.html) cannot be applied to this problem. Is that right? I am getting slightly confused. I though we can use the same principal: 1st person can be teamed up with another 6 and then another 5, so 6*5 to choose a team of 3. And then 4th person can be teamed up with another 3, while 6th person - with only 1. In total: 6*5*3...
I would much appreciate your help! Thank you!!
In the problem at the link above, the first method that I suggested works well because each person must be paired with exactly one other person.
In the problem here, the situation is a bit more complex:
Since the 7 people are being divided into two pairs and a GROUP OF 3, each person could be combined either with exactly one other person or with a PAIR OF PEOPLE.
Thus, to apply the same reasoning here, we need to consider TWO CASES.
Case 1: Person A is in the group of 3
From the 6 remaining people, 2 must be selected to be with Person A.
Number of ways to choose 2 people from 6 = 6C2 = (6*5)/(2*1) = 15.
4 people left.
The next person selected can be paired with any of the 3 remaining people:
Number of options = 3.
2 people left.
The next person selected must be paired with the 1 remaining person:
Number of options = 1.
To combine these options, we multiply:
15*3*1 = 45.
Case 2: Person A is NOT in the group of 3
From the 6 remaining people, a group of 3 must be formed.
Number of ways to choose 3 people from 6 = 6C3 = (6*5*4)/(3*2*1) = 20.
4 people left, including Person A.
Person A can be paired with any of 3 remaining people:
Number of options = 3.
2 people left.
The next person selected must be paired with the 1 remaining person:
Number of options = 1.
To combine these options, we multiply:
20*3*1 = 60.
Total ways = 45+60 = 105.
I prefer the approach suggested in my post above.
Last edited by GMATGuruNY on Wed Jan 23, 2013 8:22 pm, edited 1 time in total.
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Tommy Wallach
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Hey Guys,
I'm afraid there are some mistakes being made in these questions, as far as I can see (unless I'm misreading them somehow). Also, these methods are not the most efficient. I deeply recommend everyone move to the slots method, as opposed to attempting to use equations, which tend to break down when the questions get complicated.
Slots Method
1. Find # of slots (decision points)
2. Fill in each slot with # of choices at that decision point
3. Figure out if order matters or not
4. If order matters, simply multiply all slots
5. If order doesn't matter, multiply all slots and divide by the # of slots factorial wherever order doesn't matter. (this last step removes the duplicates when order doesn't matter)
For this question (7 people, divided into two groups of 2 and one group of 3):
1. 7 slots
2. 7 people, so:
7 6 5 4 3 2 1
3. Does order matter or not? It matters between the groups, but not within them, so:
(7*6)/2! * (5*4)/2! * (3*2*1)/3! = 21 * 10 = 210
For the other question (7 people in 2 groups, one of 3 and one of 2):
2. 7 people, so:
7 6 5 4 3
3. Does order matter or not: In this case, it matters between the two groups, but not within them, so...
(7*6)/2! * (5*4*3)/3! = 21 * 10 = 210
The answer to the questions is the same!
-t
I'm afraid there are some mistakes being made in these questions, as far as I can see (unless I'm misreading them somehow). Also, these methods are not the most efficient. I deeply recommend everyone move to the slots method, as opposed to attempting to use equations, which tend to break down when the questions get complicated.
Slots Method
1. Find # of slots (decision points)
2. Fill in each slot with # of choices at that decision point
3. Figure out if order matters or not
4. If order matters, simply multiply all slots
5. If order doesn't matter, multiply all slots and divide by the # of slots factorial wherever order doesn't matter. (this last step removes the duplicates when order doesn't matter)
For this question (7 people, divided into two groups of 2 and one group of 3):
1. 7 slots
2. 7 people, so:
7 6 5 4 3 2 1
3. Does order matter or not? It matters between the groups, but not within them, so:
(7*6)/2! * (5*4)/2! * (3*2*1)/3! = 21 * 10 = 210
For the other question (7 people in 2 groups, one of 3 and one of 2):
2. 7 people, so:
7 6 5 4 3
3. Does order matter or not: In this case, it matters between the two groups, but not within them, so...
(7*6)/2! * (5*4*3)/3! = 21 * 10 = 210
The answer to the questions is the same!
-t
Tommy Wallach, Company Expert
ManhattanGMAT
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GMATGuruNY
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This would be the correct answer if the question asked for the number of ways that the 7 people could be ASSIGNED to three DIFFERENT groups.Tommy Wallach wrote:
For this question (7 people, divided into two groups of 2 and one group of 3):
1. 7 slots
2. 7 people, so:
7 6 5 4 3 2 1
3. Does order matter or not? It matters between the groups, but not within them, so:
(7*6)/2! * (5*4)/2! * (3*2*1)/3! = 21 * 10 = 210
Let the 7 people be A, B, C, D, E, F and G.
Let say that Project One requires 3 people, Project Two requires 2 people, and Project Three requires 2 people.
The following represent two different assignments:
Project One = ABC, Project Two = DE, Project Three = FG.
Project One = ABC, Project Two = FG, Project Three = DE.
But the question above asks for the number of ways that the 7 people can be DIVIDED.
While ABC-DE-FG and ABC-FG-DE represent two different ways to ASSIGN the 7 people, ABC-DE-FG and ABC-FG-DE represent only one way to DIVIDE the 7 people.
Once a group of 3 has been chosen, the order of the two pairs doesn't matter: DE-FG and FG-DE represent the same way of dividing the 4 remaining people into pairs.
Since the order of the two pairs doesn't matter, the result above (210) must be divided by the number of ways that the two pairs can be arranged (2!):
210/2! = 105.
Check here for an official question that asks for the number of ways that 8 people can be divided into pairs:
https://www.beatthegmat.com/forming-teams-t73034.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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
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
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GMAT/MBA Expert
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Tommy Wallach
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Hey Guru et. al.,
I see the issue, I read "How many different groups can be made?" not "How many ways can they be divided into these groups?" Pays to read the original question closely! Thanks for the correction. Crisis averted!
-t
I see the issue, I read "How many different groups can be made?" not "How many ways can they be divided into these groups?" Pays to read the original question closely! Thanks for the correction. Crisis averted!
-t
Tommy Wallach, Company Expert
ManhattanGMAT
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