Can you please explain the following:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
A. 20
B. 40
C. 50
D. 80
E. 120
Thanks,
PJ
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combinations problem
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kmittal82
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Good question.
Let the couples be a1a2, b1b2, c1c2, d1d1 and e1e2 i.e. 10 people
Now, we can chose 1 out of those in 10 ways. Lets say we chose a1
For the next person, we have 9 to chose from, but we cant chose a2, hence we only really have 8 we can chose from. Lets say we picked b1
For the last place, we have 8 people remaining, out of which 2 can't be chosen to fulfill the condition, so we have 6 to chose from (we already have a1 and b1, we can't pick a2 or b2, hence 2 out of the remaining options are invalid)
Total ways of chosing = 10 x 8 x 6 = 480
Each of those 3 people can be further rearranged in 6 ways
Therefore, number of committees = 480/6 = 80
(D) should be the answer
Let the couples be a1a2, b1b2, c1c2, d1d1 and e1e2 i.e. 10 people
Now, we can chose 1 out of those in 10 ways. Lets say we chose a1
For the next person, we have 9 to chose from, but we cant chose a2, hence we only really have 8 we can chose from. Lets say we picked b1
For the last place, we have 8 people remaining, out of which 2 can't be chosen to fulfill the condition, so we have 6 to chose from (we already have a1 and b1, we can't pick a2 or b2, hence 2 out of the remaining options are invalid)
Total ways of chosing = 10 x 8 x 6 = 480
Each of those 3 people can be further rearranged in 6 ways
Therefore, number of committees = 480/6 = 80
(D) should be the answer
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bdevas01
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Note: This is a combinatorics problem; you should know what cominations and permutations are in order to solve this problem correctly.
Now, think of it like this:
Number of committees without a married couple = total number of committees possible - number of committees with a married couple.
To compute the total number of committees possible, we use the combinations formula: 10C3 = ( (10!) / (3! * 7!) ), which equals 120.
Now we need to know the number of committees, out of the 120 possibilities, that contain married couples. We know that if a committee is going to have a married couple, it has to be one of the 5 couples plus one of the 8 remaining individuals. Therefore, the number of committees with a married couple is 5 x 8, which equals 40.
Now we can plug these values into our original equation:
Number of committees without a married couple = 120 - 40.
The answer is (D) 80.
Let me know if any of this is unclear to you.
Now, think of it like this:
Number of committees without a married couple = total number of committees possible - number of committees with a married couple.
To compute the total number of committees possible, we use the combinations formula: 10C3 = ( (10!) / (3! * 7!) ), which equals 120.
Now we need to know the number of committees, out of the 120 possibilities, that contain married couples. We know that if a committee is going to have a married couple, it has to be one of the 5 couples plus one of the 8 remaining individuals. Therefore, the number of committees with a married couple is 5 x 8, which equals 40.
Now we can plug these values into our original equation:
Number of committees without a married couple = 120 - 40.
The answer is (D) 80.
Let me know if any of this is unclear to you.
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debmalya_dutta
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Hi Guys,kmittal82 wrote:Good question.
Let the couples be a1a2, b1b2, c1c2, d1d1 and e1e2 i.e. 10 people
Now, we can chose 1 out of those in 10 ways. Lets say we chose a1
For the next person, we have 9 to chose from, but we cant chose a2, hence we only really have 8 we can chose from. Lets say we picked b1
For the last place, we have 8 people remaining, out of which 2 can't be chosen to fulfill the condition, so we have 6 to chose from (we already have a1 and b1, we can't pick a2 or b2, hence 2 out of the remaining options are invalid)
Total ways of chosing = 10 x 8 x 6 = 480
Each of those 3 people can be further rearranged in 6 ways
Therefore, number of committees = 480/6 = 80
(D) should be the answer
My understanding was that the total ways of choosing = 10 x 8 x 6 = 480 does not include the 6 ways in which the 3 people can be arranged . Is that the case though ?
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bdevas01
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debmalya_dutta wrote:Debmalya,kmittal82 wrote: Hi Guys,
My understanding was that the total ways of choosing = 10 x 8 x 6 = 480 does not include the 6 ways in which the 3 people can be arranged . Is that the case though ?
Actually, the 480 DOES include the 6 ways that the 3 people can be rearranged. That's actually why kmittal divided the number by 4.
Last edited by bdevas01 on Thu Jul 15, 2010 11:40 am, edited 1 time in total.
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debmalya_dutta
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So bdevas01 ,
480 is the answer ? If the arrangement in the group would be applicable (hypothetical scenario) sense then the total number of ways would have been 480 * 6....right ?
480 is the answer ? If the arrangement in the group would be applicable (hypothetical scenario) sense then the total number of ways would have been 480 * 6....right ?
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bdevas01
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That was my mistake, I thought u were talking about something else. I edited my previous post to reflect the correct information.debmalya_dutta wrote:So bdevas01 ,
480 is the answer ? If the arrangement in the group would be applicable (hypothetical scenario) sense then the total number of ways would have been 480 * 6....right ?
You are correct in saying 480 includes the 6 ways to arrange each person so there is no need to multiply 480 * 6. Any further questions?
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sreak1089
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5 couples => 10 people.
3-member committee needs to be formed, which can be done in 10C3 ways.
However, it includes married couples. 1 married couple can be selected in 5C1 ways. Another
one person can be selected in 8C1 ways.
thus, the answer is 10C3 - 5C1 * 8C1 = 120 - 5*8 = 80.
hence d
3-member committee needs to be formed, which can be done in 10C3 ways.
However, it includes married couples. 1 married couple can be selected in 5C1 ways. Another
one person can be selected in 8C1 ways.
thus, the answer is 10C3 - 5C1 * 8C1 = 120 - 5*8 = 80.
hence d
