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

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Bunuel

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Bunuel

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Bunuel

12 Days of Christmas 2024 - 2025 Competition with $30,000 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

GMAT Club's Official Explanation:

When selecting 6 people out of 8, only one group of 6 will include all three married couples. Since we are interested in all combinations except the one that includes all three married couples, we subtract this single case. Therefore, 8C6 - 1 = 27 gives the total number of 6-member committees without all three married couples.

Answer: D.

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Updated on: 17 Dec 2024, 10:44

Originally posted by hr1212 on 16 Dec 2024, 09:10.
Last edited by hr1212 on 17 Dec 2024, 10:44, edited 1 time in total.

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Total ways to select 6 members from 3 couples and 2 bachelors (8 people) = 8C6 = 8C2 = 8*7/2 = 28

Number of ways to select all 3 couples = 1 (Select all 6 people who are couples)

Number of ways to select at most 2 couples
= Total selections - Number of ways to select all 3 couples
= 28 - 1
= 27

Answer : D

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The question mentions at most 2 couples. So firstly, the total ways of making a committee of 6 from 8 people (3 couples and 2 bachelors) is 8C6 which is same as 8C2 = 28.
Number of ways of selecting a 6 people such that all are couples = 6C6 = 1
Hence, no of committees that can be formed if the committee includes at most two married couples = 28 - 1 = 27

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We have to select 6 members from 8 people (3 couples and 2 bachelors)

At most 2 married couples = Total possible ways - Three married couples

=> 8C6 - 1
=> 28 - 1
=> 27

Answer. D

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Ans : D (IMO)

Total ways of forming a six-member team from total of 3*2 + 2 = 8 total members = 8C6 = 28
No. of ways all married couples can be in the team = 1

At most 2 couples in the team = (Total ways or number of teams - No. of ways all married couples can be in the team)
28-1 = 27

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In total we have 8 people (3 Married couples and 2 Bachelors). We are asked to make a committee of 6 people such that atmost 2 married couples can be selected. Basically, we need to subtract the case that includes 3 married couples in the committee of 6 from all possible ways of selecting 6 people from 8.

6 people can be selected from 8 in 8C6 ways, which is equal to 28. There is only 1 possibility of including all 3 married couples in committee. So the final answer is 28 - 1 = 27. Thank you.

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Selecting at most two couples from 3 .3c2 = 3 ways. Selecting 2 out of remaining 1 from couple one member and 2 remaining bachelor members in 2 ways. Therefore 5 ways

Selecing 1 couple from 3 in 3c1 = 3 ways. And 4 members from 6 members in 4 ways. 1 memeber each from other two couples and 2 from bachelors. 7 ways

Therefore 5 +7 = 12 ways

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There are three married couples (2*3 = 6 members) and 2 bachelors. So totally 8 members in the group. From these 8 members, 6 members committee needs to be formed such that it has at most 2 married couples.

"At most 2 married couples" includes 0, 1, or 2 married couples. Instead of finding these all, we can find total possible ways to choose the members, then subtract the number of ways to choose 3 married couples.

At most 2 married couples = Total ways - 3 married couples

Total number of ways ----> 6 members are selected from the group of 8. 8C6 = (8!)/(6!*2!) = 28 ways

Number of ways three married couple are being selected ---> Since we have exactly 3 married couples and 6 members need to be selected, it's only one way.

So the answer is 28-1 = 27 ways (option D)

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Answer. D)27
12+15=27 ways

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16 Dec 2024, 09:26

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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?

Total number of different committees that can be formed without any restrictions = (3*2+2)C6 = 8C6 = 8!/6!2! = 28

The number of committees with 3 married couples = 6C6 = 1

The number of different committees can be formed if the committee includes at most two married couples = 28 - 1 = 27

IMO D

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16 Dec 2024, 09:27

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Total Members = 8 (3 married couples = 6 individuals and 2 bachelors)
A committee including at most 2 married couples can be formed by = Total - all married couples included

Total committee members = 8!/6!2! = 28
All married couples = 3!/3! = 1

So, final answer = 28-1
= 27 (OPTION D)

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Bunuel

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Consider a case where all the married couple are in the committee, and the inverse to it is what the question is asking us for

So,
total of 8 people including both bachelors and couple, these must be fit into 6 spots
8C6 is the total number of ways we can select them which is equal to 8!/(8-6)!6! = 28.

And the case we are considering is all the married couple in the committee which is 6 persons in the couple and 6 spots in the committee.
So, 6C6 = 1

All the cases except this case include atmost 2 married couple in the committee.
Hence, the total number of ways we can select atmost 2 married couple in the committee of 6 spots is 28 - 1 = 27.

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The first thing to see is if the 6 member committee can be formed with 0 or 1 married couple or not. Given we need to find ways to form the committee with at most 2. 0 married couples would leave us with 2 bachelors only and 6 spots to fill and 1 married couple would leave us with 2 bachelors and 4 spots to fill. Hnece 0 and 1 married couple scenarios are not valid given then 6 member constraint. Now for the 2 married couples scenario , there are 2 spots remaining and 2 bachelors. In order to choose 2 married couples from the available 3 - We choose 1 member from the 6 people in the married group ( 3 couples = 6 people ) and then choose the 1 more member from the 4 people remaining (1 from the original selection and 1 from the partner of the original selection are removed from the 6). This can be done in 6 x 1 x 4 x 1 ways (counting principal) next we select 2 people from the 2 bachelors which can be done in 2C2 ways. which gives us : 6 x 1 x 4 x 1 x 1 = 24

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Bunuel

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Case 1: 1 couple, 1 person each from 2 couples & 1 from bachelor
= 12, 3C1*2C1*2C1*2C2

case 2: 2 couple, 1 from remaining 1 couple & 1 bachelor
= 12, 3C2*2C1*2C1

case 3: 2 Couples & 2 bachelor
= 3, 3C2*2C2

Total: 27

IMO D

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Solution

1) Identify the total number of people: There are 3 married couples, so that makes 6 individuals and there are 2 bachelors.

In total, there are 6+2=8 people.

2) Total number of ways to choose 6 out of 8: First, find the total number of ways to form a committee of 6 people from the 8 individuals without any restrictions.

8C6 = 8!/6!(8!-6!)= 28

3) Identify the restriction "at most two married couples": The only arrangement we must exclude is the one where all 3 married couples are chosen, which is exactly one combination. Therefore, the valid number of committees is 28−1=27.

Answer: Option D

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No of outcomes with atmost 2 couples= total outcomes - no of outcomes with exactly 3 couples
=8C6-1
=28-1
=27
Not sure if this is the right approach, look forward to the correct solution

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d. 27 is the answer for the question

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? 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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16 Dec 2024, 09:47

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we have two cases
case 1- 1 couple out of 3 in 3c1 and 1 1 person from next 2 couple in 2c1 ways & both singles can be selected in 1 way so 3c1*2c1*2c1*2c2

case 2- 2 couples out of 3 in 3c2 & 1 person from 1 couple in 2c1 ways & 1 out of 2 singles in 2c1 ways so 3c2*2c1*2c1

adding case 1 & case 2 =24

note there can't be the case that no couple can be selected. I marked incorrect option in first because of that mis reading

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