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

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HarshaBujji

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Bunuel

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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?

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

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To get the criteria for almost 2 couples means = #total - # 3 couples.

Total = 8C6 = 28.

If 3 couple pairs are selected only one possible selection.

Desired outcome = 28 -1 =27.

IMO D

ashminipoddar10

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Total ways in which we can form a committee of 6 from 8 people = 8c6 = 28
Total ways in which we can form a committee of 6 from 3 couples only= 1
Therefore, 27 (28-1) ways in which we can form a committee with at most 2 couples

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Ans: D. 27

3 couples and 2 bachelors

so, there are 3 ways:
1. Select 1 Married couple and then select 1 person each from the remaining 2 pairs and then select the 2 bacelors
2. Select 2 Married couple and then select 1 person from the remaining married couple and 1 from the bachelors
3. Select 2 Married couples and select the 2 bachelors
Hence,
(3C1*2C1*2C1*2C2)+(3C2*2C1*2C1)+(3C2*2C2)
=(3*2*2*1)+(3*2*2)+(3*1)
=12+12+3
=27

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NilayMaheshwari

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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?

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

At most 2 means 0, 1 or 2. We can either calculate the number of ways to form a committee for each case or we can subtract the possibility of 3 couples from the total.
Total number of committees = 8C6 = 28.
Number of committees with 3 couples= 1.
Therefore at most 2 couples = 28-1 = 27.
IMO option D

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atmost 2 married couple = 1married couple + 2married couple

But here we need to make 6 member team
so 2 married couple + 2 bachelors
3C2 X 2C2

venkatesh_sd

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B. 24

Explanation:
Condition - Max 2 married couples

There are 3 scenarios
1) 0 married couple - This is not possible.
2) 1 married couple - (12 ways = 3 * (4 * 1))
> - Ways of selecting 1 couple - 3C1 - 3
> - Ways of selecting 4 from the rest without selecting another couple
> -- ways of selecting 2 bachelors - 2C2 - 1
> -- ways of selecting 2 from the other married couple- 2C1 + 2C1 - 4

3) 2 married couple - (12 ways )
> - Ways of selecting 2 couples - 3C2 - 3
> - Ways of selecting 2 from rest without selecting another couple -
> -- 0 from the married couple -> 2C2 - 1
> -- 1 from the married couple -> 2C1 + 2C1 - 4

Since we need to form 6 member committee,

Bunuel

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Updated on: 18 Dec 2024, 09:00

Originally posted by twinkle2311 on 16 Dec 2024, 11:01.
Last edited by twinkle2311 on 18 Dec 2024, 09:00, edited 1 time in total.

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We need a 6-member committee from 3 married couples (6 people) and 2 bachelors, with at most 2 married couples.
Total ways to choose 6 people from 8:
8C6 = 28
Subtract cases with all 3 married couples:
If all 3 married couples are selected, there’s exactly 1 way:
3C3 = 1
Committees with at most 2 married couples:
28 - 1 = 27
Final Answer: 27

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8!/6!*2!

the order does not matter.

the total number of options is 28. Since there is only one combination that would include all the couples, 6, the correct answer is D, 27.

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Total number of people = 8
So 6 people out of 8 are selected, 8C6 = 8*7/2 = 28
At most this selection can have 2 married couples, so the selection that needs to be removed is the one that contains 3 married couples, which can be formed in only 1 way
So the correct answer is 28 -1 =27

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The total ways in which a committee of 6/ 8 can be formed = 8C6

The no: of ways in which a committee can be formed with atmost two couple= Total ways- ( A committee of 6 with 3 couples)

Answer: 8C6-1= 28-1 = 27

D IMO

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The answer is option D 27 as the number of committee member to be formed is 6 out of total 8 members

B denotes bachelor and MC denotes Married couple

The combination can be 2 B and 4 members from 2 MC or 1 B and 2MC and 1 from one MC and so on. But the combination cannot be
3MC for all the six members. As the condition says that there can be at most 2 MC means there can be zero, one or two MC but not more than 2.

So, 8C6 -1 = 28-1 = 27

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Hi Everyone!

I wish us a great competition

The answer is: All possibilities - only the couples chosen.
All possibilities => 6 people of committe to choose from 8 = [8*7*6*5*4*3][/6*5*4*3*2*1] = [8*7][/2] = 28
Choose only the couples => [6*5*4*3*2*1][/6*5*4*3*2*1] = 1

28-1 = 27 Our answer.

rns2812

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The question requires atmost 2 married couples - possible cases are 0 married couple, 1 married couple and 2 married couples.

Total people available = 3*2 (Married couple) + 2 Bachelors
People needed = 6

Case 1 : 0 married couple

We choose 2 bachelors and we need 4 non-married people from 3 couples. This is not possible as the 4th person will be someone's partner

Number of possibilities = 0

Case 2 : 1 married couple

We choose 1 married couple(2 people) from 3 married couples and for the rest 4 people we can choose 2 bachelors from 2 bachelors and 2 non-married people (1 partner from each couple)

Number of possibilities = $3C_{1}$ * ( $2C_{2}$ * $2C_{1}$ * $2C_{1}$ ) = 3*1*2*2 = 12

Case 3 : 2 married couples

We choose 2 married couples(2 people) from 3 married couples and for the rest 2 people we can choose following possibilities

a) 2 bachelors from 2 bachelors
b) 1 bachelor from 2 bachelors and 1 partner from the remaining couples

For a)

Number of possibilities = $3C_{2}$ * ( $2C_{2}$ ) = 3*1 = 3

For b)

Number of possibilities = $3C_{2}$ * ( $2C_{1}$ * $2C_{1}$ ) = 3*2*2 = 12

Total number of possibilities for Case 3 = 15

Total possibilities for question = Case 1 + Case 2 + Case 3 = 0 + 12 + 15 = 27 (OPTION D)

Bunuel

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Abhishek0809

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we cant select 0 couples.

So if we select 1 couple then total ways are 3 * 4 = 12

if we select 2 couple then total are 3 * 4 = 12 + 3 = 15

answer is 27

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3 married couples means 6 people in total and with 2 bachelors will make the total to 8.
so, out of 8 we have to choose 6.
we have given 2 couple at most can be presented, so 3 cases will be there-
1. case 1- 0 couples-
if we take 0 couples then we are essentially eliminating 3 people from the selection so then we will have to choose 6 from 5 people which is not possible so this case is not valid.

2. CASE 2- 1 COUPLE

selecting 1 couple from 3 is 3c1= 3
for remaining 4 spots we will pick 1 from 2 couples each which is 2c1 and 2c1= 2*2= 4
we will choose 2 from 2 bachelors remaining= 2c2=2.
TOTAL FOR THIS CASE= 3*4*1= 12

3.CASE 3- 2 COUPLE

selecting 2 couple from 3 is- 3c2= 3
now you have 4 people to choose from for 2 spots- we will have 2 cases here.
1. choose from the couple- 2c1 and 1 from the bachelors 2c1= 2*2= 4
2. add 3 as another member from the couple or the bachelors could also be chosen.=3
TOTAL FOR THIS CASE- 3*4+3=15

TOTAL COMBINATIONS= 12+15= 27

Nikhil17bhatt

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IMO D

The best way to these kind of questions is by taking the total outcomes and subtracting the worst case scenarior

Total outcomes will be 8C6 if total 8 people ( 3 Married couple ( Man, woman) and 2 Bachelor) = 28
Worst case is all 3 married couples are there in the group of 6 which is 1 hence
28-1= 27

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Since a group of 6 has to be formed and there are 6 people who are married (3 couples in total); so the number of committees that can be formed can be found by combination i.e. -

= Total number of committees from 8 people - Number of committee from all married couples
= 8C6 - 6C6
= 28 - 1
= 27

imo D

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PrateekShawarma

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Total No of ways - All couples = Would give us atmost 2 couples

8
C - 1 = 28 -1 = 27
6

Bunuel

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Bunuel

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Assume that the committee has one married couple -

Number of ways of selecting one married couple out of three married couples = 3C1 = 3
We have got two people.
We need two more.
As there are two bachelors, we must include them in the committee as no other couples can be included. From the other two married couples we need to select only one.
Number of ways = 3 * 2 * 2 = 12

Assume that the committee has two married couples -

Number of ways in which two of three couples can be selected = 3C2 = 3
Out of the four remaining people (2 bachelors and one couple), we need to select either both the bachelors or one bachelor and one of the couple
Number of ways = 3 * (1+2*2) = 3*5 = 15

Total = 15+ 12 = 27

Option D

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||
D. 27.

The Dinner Party Riddle
Picture this: You’re at the world’s most exclusive soirée, and everyone’s whispering, Who’s on the committee? The bachelors are suave, the couples are classy, but there’s a catch, no more than two couples can pair up.

You calculate it like a maestro arranging a perfect symphony:
No couples: Just a mix of solo acts.
One couple: A balanced duet amidst the solos.
Two couples: Enough romance, but not too much.
When you tally up every sparkling possibility, the magic number is 27—a party where the drama is just right, and everyone gets their moment in the spotlight.
||

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