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CasperMonday
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GMATPrep: Combinatorics, easy one 31 Aug 2009, 06:19
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a. 16
b. 24
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e. 32

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GMATPrep: Combinatorics, easy one 31 Aug 2009, 06:35
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i have also got this ques in which probability was asked...


See , there are 4 coupples select any three : 4c3

now from each couple select one : 2c1

so 4c3 * 2c1* 2c1* 2c1 =32 ans.....Hope it helps....
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GMATPrep: Combinatorics, easy one 31 Aug 2009, 16:01
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The total number of possible committees is \(8C3=\frac{8!}{5!\times 3!}=56\)
The total number of committees including a couple is \(4\times 6=24\). A commite can be formed of a couple plus any other of the remaining 6.
So the possible committees that do not include a couple is \(56-24=32\).
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GMATPrep: Combinatorics, easy one 31 Aug 2009, 22:36
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:( cant see image can some one repost question
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GMATPrep: Combinatorics, easy one 01 Sep 2009, 08:23
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LenaA
The total number of possible committees is \(8C3=\frac{8!}{5!\times 3!}=56\)
The total number of committees including a couple is \(4\times 6=24\). A commite can be formed of a couple plus any other of the remaining 6.
So the possible committees that do not include a couple is \(56-24=32\).
Show more


Hi Lena,

can you please tell me how did u get this part ,total number of comittees including a couple 4*6=24

the rest part is clear.

Thanks in advance
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GMATPrep: Combinatorics, easy one 01 Sep 2009, 11:05
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gurpreet07
LenaA
The total number of possible committees is \(8C3=\frac{8!}{5!\times 3!}=56\)
The total number of committees including a couple is \(4\times 6=24\). A commite can be formed of a couple plus any other of the remaining 6.
So the possible committees that do not include a couple is \(56-24=32\).


Hi Lena,

can you please tell me how did u get this part ,total number of comittees including a couple 4*6=24

the rest part is clear.

Thanks in advance
Show more

Consider a particular couple. You want to form a committee that will include this couple. How many committees like that can be formed? Since there are 3 members in the committee, we need to add 1 more person to this couple. In total there are 8 people, minus this couple, six people are left from which you can choose an additional member. So in total you have 6 options to form a committee consisting of a particular couple and an additional member.
But you have 4 couples. Hence, the the total number of commitees including any couple will be \(4\times 6=24\)
Hope this helps
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GMATPrep: Combinatorics, easy one 01 Sep 2009, 11:19
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LenaA
gurpreet07
LenaA
The total number of possible committees is \(8C3=\frac{8!}{5!\times 3!}=56\)
The total number of committees including a couple is \(4\times 6=24\). A commite can be formed of a couple plus any other of the remaining 6.
So the possible committees that do not include a couple is \(56-24=32\).


Hi Lena,

can you please tell me how did u get this part ,total number of comittees including a couple 4*6=24

the rest part is clear.

Thanks in advance

Consider a particular couple. You want to form a committee that will include this couple. How many committees like that can be formed? Since there are 3 members in the committee, we need to add 1 more person to this couple. In total there are 8 people, minus this couple, six people are left from which you can choose an additional member. So in total you have 6 options to form a committee consisting of a particular couple and an additional member.
But you have 4 couples. Hence, the the total number of commitees including any couple will be \(4\times 6=24\)
Hope this helps
Show more


Thanks for the response....I got it now



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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