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12 Dec 2024, 02:00
Kudos
Bookmarks
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:)
16 Dec 2025, 04:57
Kudos
Bookmarks
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