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12 Days of Christmas GMAT Competition - Day 1: From a group consisting

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Bunuel

12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

From a group consisting of 3 married couples and 2 bachelors, a committee of 6 members is to be formed. How many different committees can be formed if the committee includes at most two married couples?

A. 12
B. 24
C. 25
D. 27
E. 28

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3 pairs and 2 bachelors
Task = 6 members at most 2 pairs = Total - The number of ways of having more than 2 pairs (3 pairs) = 8C6 - 1 = 8x7/2 - 1 = 27
Correct answer = 27 --> D

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Atmost 2 married couples => 0 couples, 1 couple, 2 couples

Non-event case: Total - 3 couples case = 8C6 - (6c6) {choose 6 of 6 couple members, and no bachelors} = ((8*7)/2) - 1 = 28-1 = 27

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We have 3 Married couples --> 6 people
And 2 bachelors

2 scenarios
1) 2 Married couple + 1 Bachelor + 1 member of remaining couple (OR) 2 Married Couple + 2 Bachelor
2) 1 Married couple + 2 bachelors + 2 oppposite couples

1) 3C2 * 2C1 * 2C1 + 3C2 * 2C2 = 15
2) 3C1 * 2C2 * 2C1 * 2C1 = 12

Total = 27 ways

Alternate :
Total no. of ways - 3 Married Couple
8C6 - 3C3 = 28 - 1 = 27

Bunuel

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Bunuel

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Used the negation

Atmost 2 married couples = All - 3 married couples
= 8C6 - 1 ( as 3 MC = 6 people committee)
= 28-1 = 27

IMO D

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0 couples= 0
1 couple= 3C1x 4=12 ways
2 couples=3C2x(4C2-1)=15 ways
total=27 ways

or
Total - 3 married couples= 8C6-1=28-1=27

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Given:
Total people = 8 (3 married couples + 2 bachelors).
We need to form a committee of 6.

Step 1: Total committees that can be formed (A):
Total ways to pick 6 from 8 = 8C6 = 28

Step 2: Committees with all 3 married couples (B):
If all 3 married couples are included (i.e. 6 people), only 1 such committee can be formed.

Step 3: Different committees with at most 2 couples
= Total committees - Committees with all 3 couples
= 28 - 1
= 27

Answer: 27 committees.

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We have 3 couples, and 2 bachelors = a total of 8 people, and need to form a committee of 6 members.
The problem can be broken into 3 different cases.
Case 1: No Married Couple Included:

This is impossible since selecting one person from each couple, and both bachelor's would lead to only 5 members in the committee.
Case 2: One Married Couple Included:

We can select a couple in 3C1 ways, and a person from remaining 2 couple in 2*2C1 ways.
count = 3 * 2*2 = 12 ways.
Case 3: Two Married Couples Included:

selecting 2 married couples out of 3 couples (4 people) = 3C2
Remaining 2 people here can be selected in 2 different ways.
1. Select both the bachelors = 1 way
2. Select one person from 3rd couple, and one bachelor = 2*2 = 4
Total = 3 * (1+4) = 15 ways

Adding all 0 + 12 + 15 = 27 ways.
Ans D

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3 married couple
2 bachelors

2 married couple = 3c2 x 1 = 3
1 married couple = 3c1 x 4 x 1= 12

2 married couple and 1 bachelor = 3c2 x 4 = 12
therefore total = 3 + 12 + 12 = 27

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Bunuel

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8!/6! = 28 less 3 married couple committee (1) = 27 (D)

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In combination problems and PS problems in general it is best if we figure out a method before jumping onto the solution, and also one has to be careful if the problem entails only selection or both selection and arrangement.
Here, we have to select 6 people out of: 2 bachelors and 3 married couples.

The number of ways of selecting any 6 people out of 8 is 8*7*6! /6!*2! = 28
the people will get selected in various ways:-

1. Three couples
2. Two couples, one from 3rd couple and one bachelor
3. One couple, two from two different couples and two bachelors
4. Two couples and two bachelors

Now at this stage we can apply two methods, either we can count the individual cases or we can simply subtract case 1 from total of 28 as out of all these cases, only the first one, which includes three couples, has more than two married couple.
Since GMAT requires us to find the most efficient way to answer a question, let’s go via subtraction;

How many different committees can be formed such that there are 3 couples?
It will be equal to 3 choose 3 i.e. 1 way of selection.

Number of different committees of 6 people such that there are at most two married couple = 28-1=27.
ANSWER IS D.

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Total 8 people .. Select any 6 in 8C6 ways = 28
One combination will have all 3 couples, occupying all 6 positions. Subtract this combination.

Required combination = 28 - 1 = 27

Ans D

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There can only be two cases here :

Case 1: With two married couples

2 married couples can be chosen in 3C2 ways. Two more people can be chosen in two ways - One from among the remaining married couple and one from the bachelors (or) both bachelors can be selected.

In the first case, one person from among the remaining married couple can be chosen in 2 ways. One from among the bachelors can be chosen in 2 ways

In the second case, 2 bachelors can be chosen in 1 way.

Total combinations = 3c1*2*2 + 3C1*1 = 15

Case 2: With one married couple

1 married couple can be chosen in 3c1 ways. From the remaining two couples, I can only chose a maximum of 2 people without forming a married couple. The rest 2 people to be selected are necessarily bachelors

Total combinations = 3C2* 4*1 = 12

Total ways = 15+12 =27

Hence, Option D

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we are asked about at most 2 married couples (MC)
So we can calculate this as = Total - (cases with >2 MC)
=> total - cases where 3 MC
total cases = 8C6 {3 MC = 6 people, plus 2 B}
case where 3 MC =1 {all 3 MC selected, all the seats filled}

Hence 8C6-1=27

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Number of ways to select a group of 6 members out of 8 is 8C6-8*7/2=28

Out of that the number of ways we select all 3 couples will be just 1.

Hence there are 28-1=27 ways to choose at most 2 couples. Hence the answer is option (D) 27

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Ans is E
3 couples, 2 bachelors, selecting 6 people
2 couples:-
3C2*2C2=3
1 couple:-
3C1*4C1*2C1*2C2 = 24
0 couple:-
3C3=1

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Bunuel

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There will be two cases here
Case 1 :
One Married couple
choosing married couple -> 3C1 X Other couple and bachelore can be arranged in 5 ways since only one can come from each couple and one from bachelor or both from bachelor - > 3X4 = 12

Case 2 :
Two married couple -> 3C2 X one from the couple and one from bachelor or both from bachelor - > 3X5 = 15

Case 1 + case 2 = 27

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The answer is 27.
We have total 8 members (3 couple + 2 Bachelor). Of these 8 people we are tasked to select 6-member committee.

So number of ways in which we can select any 6 members out of 8 is 8c6.

How many different committees can be formed such that there are 3 couples? Only one since we have only 3 couples. We will have to select all the couples, and we will get 6 people.

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Bunuel

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Since there are only 3 couples in consideration, cases with at most 2 couples basically means: all the cases - cases with 3 couples.

Total number of cases can be calculated as follows:
Total number of people = 3*2 + 2 = 6 + 2 = 8
Number of people to be selected = 6
Number of ways = 8C6 = 28

Number of cases with 3 couples:
Total number of people = 8 (as calculated above)
Number of people to be selected = 6
Number of people already selected (as we have already selected 3 couples) = 6
Number of people that remain to be selected = 0
Number of ways = 8C0 = 1

=> Number of committees with at most 2 couples = 28 - 1 = 27

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case1-> 3c1x4= no. of ways of choosing 1 couple, 2 singles and 2 of 1 person from each of other two unselected couples
cas2-> 3c2x2x2= no. of ways of selection 2 couples with 1 of the person of the remaining couple and 1 of the person of the 2 singles.
plus, 3c2= no. of ways of selecting 2 couples and the 2 singles. therefore, 3c1x4 + 3c2x2x2 + 3c2= 27