I28-53

Question

A certain company is forming a committee of  n  people to be chosen from  n  married couples, where  n  is a two-digit even number.
 
Select for  No married couples  the number of different committees that can be formed if no two people who are married to each other are allowed to serve on the committee, and select for  Only married couples  the number of different committees that can be formed if the committee must consist only of married couples. Make only two selections, one in each column.

Bunuel · Answer

Official Solution:

Since the committee must not include any married couples, only one "representative" from each married couple can be selected. We are forming an  n -person committee, so we must select  n  married couples that will each send one "representative" to the committee. The number of ways to choose  n  couples out of  n  is  C(n, n) = 1 .
 
For each of these  n  couples, there are two choices for the "representative": either husband or wife. This results in 2 * 2 * ... * 2 ( n  times)  = 2^n  possibilities.
 
Therefore, the total number of ways to form the committee is: 
   
 C(n, n) * 2^n = 2^n .
   
As for the committees that can be formed if the committee must consist only of married couples, we are choosing  \frac{n}{2}  couples (which gives us  n  people) from  n  married couples. This is given by:
   
 C(n, \frac{n}{2}) = 
 
 = \frac{n!}{\frac{n}{2}! * (n - \frac{ n}{2})!} = 
 
 = \frac{n!}{(\frac{n}{2}!)^2}

Correct answer:

No married couples  " 2^n "

Only married couples  " \frac{n!}{(\frac{n}{2}!)^2} "

ArmaanBajaj · Answer

I don’t quite agree with the solution. Logic not fully clear

Bunuel · Answer

The solution is fully correct, precise, and detailed. If you have any specific questions, feel free to ask.

anushree01 · Answer

I like the solution - it’s helpful.

findingmyself · Answer

I did not quite understand the solution. There are 2N memebers and we need to select N members who will not have any single married pair. The first member can be selected in 2N ways, Second member in (2N-2) ways and third in 2N-4 ways and Nth member in 2N-N or N ways. Why is this approach Incorrect

For eg: I have 10 married couples from which I am selecting 10 people who are not married to each other. 1st person can be selected in 20 ways (since we have 10*2 total people), second person can be selected in 18 ways (Since I cannot select the person who is the husband/wife of the first person selected), 3rd person I can select in 16 ways, 4th in 14, 5th in 12 .....10th in 2 ways. So my total will be 20*18*16....2 Which is 2^10 (N!)
My answer hence comes to 2^N *N!

Bunuel · Answer

Please check alternative solutions here: https://gmatclub.com/forum/gmat-club-ol ... 47398.html Hope this helps.

No married couples	Only married couples
		\(n!\)
		\(n^2\)
		\(2^n\)
		\(\frac{n!}{2(n-2)!}\)
		\(\frac{n!}{(\frac{n}{2}!)^2}\)
		\(\frac{(2n)!}{(n!)^2}\)

No married couples	Only married couples
		\(n!\)
		\(n^2\)
		\(2^n\)
		\(\frac{n!}{2(n-2)!}\)
		\(\frac{n!}{(\frac{n}{2}!)^2}\)
		\(\frac{(2n)!}{(n!)^2}\)

I28-53

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