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

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Jayesh20201

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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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In the given question, the selections can be made in 8C6 ways (28 ways). The number of selections in which all couples will be in a team will be 1 since 6C6 = 1. Even logically, it does not matter how they are selected as long as they are selected. Therefore the answer is 28-1+27

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Bunuel

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8C6 Total ways 28
28 - 3 (not possible) = 25 ways

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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?

Solution:
Total Members = 8 (3 married couples = 6 individuals and 2 bachelors)
Total ways to select 6 people out of 8= 8C6= (8*7*6*5*4*3*2)/(2)(6*5*4*3*2)
=4*7=28
At most 2 married couples means: At least 1 married couple or 2 married couple or No married couple = Not (All married couples)
Committee includes at most two married couples=Total- All 3 married couples
=28-1=27

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At most two married couples --> #Total outcomes - #Outcomes with all 3 married couples

#Total outcomes : 8C6 = 28

#Outcomes with all 3 married couples : Since the committee has only 6 members, there is only one outcome including all 3 couples.

28 - 1 = 27

Ans : D

Bunuel

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3 Married Couples and 2 Bachelors - Total 3*2 + 2 = 8 people .

We are asked to include at most 2 married couples(2 Couples , 1 couple or none - Favorable Case) . Therefore ,3 Couples included- Unfavorable Case.

So , we can solve it by Favorable = Total - Unfavorable
Total is select 6 out of 8 = 8C6= 28 and Unfavorable = 6C6(All 3 Married couples) = 1
Therefore, Total - Unfavorable = 28-1=27

Bunuel

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Rex885

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Total ways to select 6 ppl from 3 couples and 2 bachelors (8 ppl) = 8C6 = 28

# of ways to select all 3 couples = 1 (Select all 6 people who are couples and no bachelors)

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

HansikaSachdeva

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There are 8 members, with 3 couples and 2 bachelors

At most 2 married couples = Total number of ways - The ways in which all married couples are included

Total number of ways = 8C6 = 8!/6!2! = 28

The ways in which all married couples are included = 6C6 = 1

Answer = 28-1 = 27

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19 Dec 2024, 08:02

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At most two married couples = all the cases minus case with all three couples.
Since there is only one case with all married couples,
answer is 8C6 - 1 that is 28-1=27

Bunuel

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Bunuel

This question was provided by GMAT Club
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What is asked?

We need to form a committee of 6 members from a group of 8 people (3 married couples and 2 bachelors). The condition is:
The committee must not include all 3 married couples.

What needs to be done?

We calculate the total number of ways to select 6 people out of 8, then subtract the single case where all 3 married couples are included.

Step-by-step thinking

Total ways to select 6 people from 8:
From 8 people, the total number of ways to form a group of 6 is:
8 c 6 = 28
Remove the case where all 3 married couples are included:
If all 3 married couples are included, that makes exactly 6 people.
There is only 1 way to select all 3 married couples.
Subtract this single case:
To exclude the committee with all 3 married couples:
28−1=27

ANSWER IS D.

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there are totally 8 members 2x 3 (married couples), 2 (bachelors).
Hence for commitee of 6 total possible selections are 8C6
Total number of selections having greater than 2 married couples = 1

Hence subracting it from total possible selections answer is D. 27

Bunuel

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chloepham

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So we have to choose R options out of total N options, which is 6 members out of 8.
We have the formula: [N!][/R!x(N-R)!] = 8! / 6!(8-6)! = 28.
However, we have 1 option where all 3 married couples are chosen, so the final answer is 28-1 = 7

==> D

Bunuel

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BillyButcher0956

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Ways to select at most 2 married couples = Total ways to select - Ways to select 3 married couples
=> 8C6 - 1 = 27 ways

Bunuel

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Yaswanth9696

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Total members: 8 (3 married couples = 6 persons and 2 bachelors).
A committee with a maximum of two married couples can be created by = Total - all married couples included.

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

The final result is 28-1 = 27 (OPTION D).

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