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GMAT Club Olympics 2025 (Day 6): A certain company is forming

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Dav2000

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Bunuel

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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.

given: n people to be chosen from n married couples.
so the total number of people = 2n.

Total probability of choosing committee unconditioned = 2nCn = (2n)!/(n!)(n!)

Probability of choosing no married couples = select one partner in committee the other one cannot be consider and the remaining number of combinations
= (2n-2)C(n-2) So not sure if it is correct explanation

Only married couples = pair of 2 formed from n couples results in n/2 groups and people in committee is n then teh probablity is given by
nC(n/2) = (n!)/((n/2)!^2) Option E.

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From the problem statement, it is evident that a committee of N individuals is to be formed from N married couples, indicating a total population of 2N persons.

Considering the constraint of only married couples:
Given N pairs of spouses, selecting N/2 pairs would suffice to form a committee of N individuals.
Thus, the number of ways to choose N/2 pairs from N is given by the combination formula:
C(N, N/2) = N! / [(N/2)! * (N/2)!]

Regarding the scenario with no married couples:
To ensure that no married couples are included, it is necessary to separate the spouses completely.
This implies that for each couple, only one member can be selected to join the committee.
In other words, for each couple, there are two possible choices (either the husband or the wife).
Given that there are N couples, the total number of possible configurations is:
2^N

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1. Non Married couples:
here we have n couples and from each couple we need exactly one member ( 2c1 )
and similarly from each of n couples we take 1 each --> (2c1)*(2c1)*(2c1) ..... n times ---> 2^n

2. Married couples:
if you select 1 couple 2 members are selected, so to get n members, we need to select n/2 couples.
nc(n/2) --> n!/((n-n/2)! * (n/2!)) --> n!/((n/2)!)^2

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07 Jul 2025, 10:32

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There are n couples => 2n people (n wives and n husbands)

For answering No married couples,
We have 2 choices among n wives and n husbands for n members of committee. This can be done in 2^n ways

For Only married couples,
We choose n/2 out of n couples to make a committee of n members.
This can be done in nC(n/2) ways
=>$ \frac{n!}{((n/2)!)^2} $

Hence answers are:

No married couples : $2^n$
Only married couples: $ \frac{n!}{((n/2)!)^2} $

Bunuel

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Firstly, observe that we have "n" couples, so "2n" people to choose from; and we need to for a committee of "n" people

No married couples:

The first person can be chosen in 2n ways
The next person can be chosen in 2n-2 ways (1 person has already been selected and his/her spouse cannot be selected)
The next person, similarly, can be chosen in 2n-4 ways

The total number of ways are: (2n)(2n-2)(2n-4)...
But we need to account for repetition as well, since we are choosing from the same pool of "2n" people. Since we are selecting n people out of them, we divide this by n!

So we get: $\frac{(2n)(2n-2)(2n-4)...}{n!}$
Taking 2 common from the numerator of each term, we get:

$\frac{2^n(n-1)(n-2).....}{n!}$

Notice how the expression after $2^n$, is nothing but n!; so that cancels out giving us:

Total number of ways for selecting no married couples as $2^n$

Married couples only:

We know that $n$ couples have $2n$ people
So, $\frac{n}{2}$ couples will have $n$ people

Since, we need to select$n$ people for the final arrangement and all of them have to be married, we can just select $\frac{n}{2}$ couples from the $n$ couples that we have and that will give us $n$ people who are married to each other

We can do that by:$ C^n_\frac{n}{2}$

which is nothing but: $\frac{n!}{(n/2)! (n/2)!}$

Answer:

No married couples: $2^n$

Only married couples: $\frac{n!}{(n/2)! (n/2)!}$

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07 Jul 2025, 10:42

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There are n couples, and you must pick one person from each couple to form the committee of n people. This avoids having any married couple in the committee.
For each couple, there are 2 choices (husband or wife).
Since there are n couples, the total number of possible selections is: 2^n

Since each couple brings 2 people, and the committee must have n people total, you must select n/2 couples.
The number of ways to choose n/2 couples from n couples is a standard combination:
(n/2) = n!/((n/2)!*(n/2)!) = n!/((n/2)!)^2

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Married couples = N
Total number of people = 2N

Goal: have X people so that no two people are from the same couple (one person from each couple)
For each couple we have to do P1 or P2 in the couple - two choices.
For N couples that would be 2^N

For no married couples we want 2^N

Goal: we want a commit of n people where everyone is forms a couple.
we want n/2 (everyone is in a married pair)
We want to choose (n/2) our of N couples. So we want (N!)/(N/2)^2

No Married : 2^N
Only Married: (N!)/(N/2 !)^2

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Total Couples: n

no married couple: committee of n people with no married couple => so you can chose either one from each couple.
No of way you can choose one out of two = 2
doing same with all the couples = 2*2*..... n times => $2^n$

married couple: Committee of n people with only married couples => you have to choose half the couples out of n couples.

Number of ways of choosing r out of n = nCr,
so choosing half out of n = $\frac{n! }{(n!/2)^2}$

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N couples then Total = 2n ( N men and N women )

Non-married couple, out of 2n we have to choose n, then from each couple we have 2 option to choose

Then for n couples each with 2 choices = 2^n

For only married couples, there are a total n married couples, 2 choose n people, we have to choose half of the couplse n/2

then by formuaa = nC(n/2) = n!/(n/2)! (n-(n/2))! = n!/(n/2)^2!

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No married couples : for committee of n members for each position any one out of the two person who are married to each other could be chosen, so that no two people who are married to each other serve on the committee. For each position we have two options so there will be 2^n ways to form the committee.

Only married couples : n/2 couples out of n couples should be chosen nCn/2 = n!/(n/2)!

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Bunuel

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Let's assume that we have 6 couples and 12 people involved.

Number of ways to select 6 people so that no couples are involved =

12 * 10 * 8 * 6 * 4 * 2 / (6 * 5 * 4 * 3 * 2) = 64

Number of ways to select 6 people so that only couples are involved = 6C3 = 6 * 5 * 4 / 3 * 2 = 20

Options

n! = 6 * 5 * 4 * 3 * 2 *1 = 60 * 12
n^2 = 36
2^6 = 8 * 8 = 64
6!/2*4! = 6 * 5 * 4!/2*4! = 15
6!/3!^2 = 6 * 5 * 4 * 3!/3! * 3! = 20
12!/6!^2 = 12 * 11 * .. = We don't need to compute as none of the answer is a multiple of 11.

Hence -

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

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Bunuel

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We need to form a committee of n people to be chosen from n married couples. Where, n is a two digit even number.

n people is chosen from a population lot of 2n peoples. ( n married couples = 2n peoples )

Lets for time being assume n = 10 , so we need to choose 10 people from 2n = 20 people.

Only married couple:

To choose 10 people we need 5 couples. Hence, we need to select 5 couples from the 10 couples.

10 C 5 = 10*9*8*7*6 / (5*4*3*2*1) = n! / [(n/2)! ]^2

= 10! / [ (10/2)! ]^2

= 10*9*8*7*6*5*4*3*2*1 / (5! * 5!)

= 10*9*8*7*6*5! / (5!* 5!)

= 10*9*8*7*6 / 5!

= 10 C 5.

Option E

No married couple:

Out of the 10 couples ( men and women) we have two choices to choose from them. So, for each couple 2 choices exist.

Hence, for 10 couples = 2*2*2*2*2*2*2*2*2*2 = 2^10

= 2^n

Option C

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No married couples:
You must form a committee of n people from n married couples, with no spouses serving together.
That means you pick exactly one person from each n couples, you have 2 choices.
Thus, the total number of ways is: $2^n$

Only married couples:
Here, the entire committee of n people must consist only of complete couples.
Since each couple contributes 2 people, you need n/2 couples total, chosen from the n available.
So the count is simply: $nC(n/2)$ = $n!/(n/2)!^2$

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For "No married couples," each of the n couples provides two independent choices (husband or wife), resulting in 2^n
possible committees. For "Only married couples," the committee of n people must consist of n/2 complete couples, so the number of ways is choosing n/2 couples from the n available couples, which is given by the combination formula n!/((n/2)!)^2.

Regards,
Lucas

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Option C and Option E are the correct answers.

Lets understand the question before we start solving for it.

The question tells us that a committee needs to be formed of n people for which will be chosen from n example: a committed of two people to be chosen from two couples married couples. Then it further adds that n is a two digit even integer.

Now the questions asks us to "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". Now lets try solving the question.

So for the first part question in which it asks us "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.

The answer to this will be 2^(n) i.e Option C because for each positive we will be having 2 options like for the first position at the committee we could either select the husband or the wife for that position, same for the second and so on. It's a bit confusing to lets try to understand it with an example.
Example: A committee of 2 people needs to be selected from 2 married couples, such that no two people are married. Now lets take couple-1 to be (x and y) and couple-2 to be (a and b). Lets see with the by just casually counting and without using any formula or property to check how many such pairs can be formed. So the pairs will be: (x and a) or (x and b) or (y and a) or (y and b) so the committee can any of these four scenarios. Similarly as discussed above if we solve this by using the formula then it can be solved like: 2*2 => 2^2 => 4.
So as we can see from this example whenever we are asked of this type of scenario instead of manually calculating the pairs we could just do 2^(number of positions in committee) to find the answer.

Now lets solve for the second question which asks us "Select for Only married couples the number of different committees that can be formed if the committee must consist only of married couples".

To get the answer for this question we just have to assume one simple thing then this question will be easily solved like any other Permutation & Combination question. So while solving this we just need to assume each couple as a single unit only which would mean that if their are 20 people i.e. 10 couples then while solving the question we need to consider them as 10 people only instead of 20, we are assume this because of the condition mentioned in the question which says that "Committee must consist of both Husband and Wife or Both Husband and Wife will be in the committee".
So the answer to this question will be Option E i.e. n!/(n!/2)^(2).

So from here we can say that Option C i.e. 2^(n) & Option E i.e. n!/(n!/2)^(2) are our answers.

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Bunuel

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Let the couples be

A1A2

B1B2

C1C2

.
.
.

we have n married couple and the committee consists of n people , hence from each married couple we can choose one. There are two ways from selecting one person from each couple

Hence the number of ways = 2 * 2 * 2 * 2... n times = 2^n

If the committee only has to have married couples then, one couple take two positions in the committee. Hence, we need n/2 married couples.

Number of ways of selecting n/2 married couples from n married couples = n C (n/2)

= n ! / (n - n/2) ! * (n/2)!

= n ! / (n/2)! ^ 2

Final Answers

1. No married couples = 2^n

2. Only Married Couples = n ! / (n/2)! ^ 2

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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.

Committee size = n
Amount of married couples = n
Therefore, amount of people = 2n

n is an even integer, greater or equal than 10 and lower than 100.

No married couples

We will have "n" couples to select "n" elements, with the restriction of not getting 2 person from the same couple.

nCn = n!/n! = 1. So, we have only one way to select couples: getting one person per couple. From each couple, we will have 2 options.

So, the different committees amount will be:
Couples-level = 1
Inside-couples = 2. Inside-couples per n couples = 2*2*...*2 n times. Therefore answer = 2^n

Only married couples

We have n spots and n couples. Since we have to bring the whole couple, we should select n/2 couples.

Combinatorics: nC(n/2) = (n!)/((n/2)!(n/2)!) = (n!) / ((n/2)!)^2 Therefore answer = (n!) / ((n/2)!)^2

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2n people, n couples. Committee of n is to be chosen.

Ways to choose n people such that only married couples are chosen:
Choose n/2 couples from n couples. This implies nC(n/2). = n!/((n/2)!)^2

Ways to choose n people such that no married couples are chosen:
There are n husbands and n wives.
nC1*(n-1Cn-1) + nc2*(n-2Cn-2) ..... nCn*1
= nC1 + nC2 + ....nCn = 2^n

No married couples: 2^n
Only married couples: n!/((n/2)!)^2

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No married couple:
If we want to have a committee of n people, with no married couple, from a set of n married couples, it means that we have to have one person from each couple, so we can have n people. The different option here is that from each couple, we have to choose 1 of the 2, and we have to do this n times, for all the couples to have n members. So the whole options are $2^n$

Only married couples:
In this case, we can assume each couple, as one option, so we need to choose $\frac{n}{2}$ between n options.
The result of this is equal to: $\frac{n!}{ { (n/2)!*(n/2)!}}$ , which is: $\frac{n!}{{(n/2)!^2} }$

n!(n2!)2

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07 Jul 2025, 16:08

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Bunuel

This question was provided by GMAT Club
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Difficult one.
My reasoning behind that is,

None-Married-Combinations:
We have
n spaces and want do fill these with n non married individuals. We view them as buckets.
First place: 2 options, second, 2 options, third 2 options,...., n-1 two options, n: two options.

Therefore, we have 2^n

For the married one:
We have space for n/2 couples.
Therefore, we look for the combinations of n/2 couples, that fit in the n spaces.
That leads us to n!/((n/2)!(n-(n/2)) -> That equals n!/((n/2)!)

Therefore, are the results:

2^n and n!/((n/2)!)