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12 Dec 2024, 02:00
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One should go through all three explanations listed here for proper understanding of sum and then ofc go ahead with the fastest one
Explanation 1
Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send only one "representatives" to the committee in \(C^4_6\) # of ways.
But these 4 chosen couples can send two persons (either husband or wife): \(2*2*2*2=2^4\).
So # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is: \(C^4_6*2^4\).
Total # of ways to choose 4 people out of 12 is \(C^4_{12}\).
\(P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\)
Answer: D.
Explanation 2
(12*10*8*6)/4!
Explanation 3
https://gmatclub.com/forum/if-4-people- ... 55-20.html
Hope it helps everyone:)
Explanation 1
Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send only one "representatives" to the committee in \(C^4_6\) # of ways.
But these 4 chosen couples can send two persons (either husband or wife): \(2*2*2*2=2^4\).
So # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is: \(C^4_6*2^4\).
Total # of ways to choose 4 people out of 12 is \(C^4_{12}\).
\(P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\)
Answer: D.
Explanation 2
(12*10*8*6)/4!
Explanation 3
https://gmatclub.com/forum/if-4-people- ... 55-20.html
Hope it helps everyone:)
28 Jun 2026, 19:10
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So basically we are multiplying by 2^4 to consider that for each couple- 1 of the two is being considered at a time right?
In case this clause of no 2 member from a couple would have been there, we would not have multiplied by 2^4 then?
In case this clause of no 2 member from a couple would have been there, we would not have multiplied by 2^4 then?
Bunuel
28 Jun 2026, 22:01
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FreshStove
Yes, exactly. The factor 2^4 is used because, after choosing 4 couples, we choose 1 person from each of those 4 couples.
29 Jun 2026, 02:47
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bibha
There are Total 12 people and 6 couples say (A1, A2), (B1, B2), (C1, C2), (D1, D2), (E1, E2)
1st person can be anyone from these 12 people so 1st person can be selected in 12 ways.
Npw only 11 persons are left and 2nd person can be anyone except for the partner of 1st person. so there are 10 ways.
Now there are 10 persons left and 3rd person can be anyone except the partner of 1st & 2nd person. there are 8 ways.
Now there are 9 persons left and 4th person can be anyone except the partners of 1st, 2nd and 3rd persons. It can be done in 6 ways.
Total number of ways of selecting 4 persons such that their partners are not in the group = 12*10*8 *6.
Total number of ways of selecting 4 persons without any restriction = 12*11*10*9
So probability that none of them would be married to each other, if 4 persons are selcted from group of 6 married couples = \( \frac{12*10*8*6}{12*11*10*9} = \frac{16}{33}\)
