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

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Elite097

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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

Total committes= 8C2= 28
Committes with all three married couples=1
Committees with at most 2 married couples= 28-1=27
Ans D

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The correct answer is D.

This is a Combination problem. We have to solve the following equation: (At most 2) = (Total) - (Exactly 3)

(Total)
We have 3 couples and 2 bachelors. In total, 8 people.
8C6 = 8C2 = 28

(Exactly 3)
6C6 = 1

(At most 2)
28 - 1 = 27

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

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

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Total people 3x2+2=8

No restrictions 8C6=28

We have to substract the number of combinations when we choose the three married couples, which is 1.

Total 28-1=27

IMO D

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First step to solving this problem is to acknowledge the "at least" words in the question.

This type of questions are often easier to solve "the other way around". In this case what the addition of "at least" means is the total amount of ways of choosing 1 or 2 couples for the committee among 3 couples and 2 bachelors (8 people).

First, calculate the total amount of ways of arranging 8 people, choosing 6 for the committee and leaving 2 outside: 8!/6!2! = 28 total ways. This would be the answer if no restrictions were added.
Second, calculate the total number of ways of choosing the 3 couples for the committee: 3!/3! = 1 there is only one way of arranging them.
Subtract the number of ways of choosing the 3 couples for the committee from the total amount of ways of selecting 6 people among 8: 28 - 1 = 27 (D)

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Answer is D) 27.

It's my first time commenting so I hope I am doing it correctly. So If we pick 6 out of 8 without restriction the answer is 28 ways (number of ways method). There we have to subtract the combination where we have picked all 3 couples out of 3 couples (3 out of 3 = 1)

28 - 1 = 27

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Answer= 27

As asked at most 2 couples, lets consider

All the possible ways for selecting 6 members from grp of 8 (3couples + 2 bachelors)= 8C6 = 28
Number of ways more than 2 couples can be chosen (Only case all 3 couples are chosen from the 3)= 3C3= 1
Therefore, At most 2 couples chosen becomes 28-1= 27.

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Bunuel

This question was provided by GMAT Club
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Let's divide the problem into multiple cases:

Case 1: 2 married couples are present in the committee.

We have 3 married couples and 2 couples to choose from, hence number of possibilities are: 3C2 => 3
The remaining two places could be as follows: (one married person, bachelor) or both bachelors only => 4 + 1 => 5
Hence total number of combination is 3*5 => 15

Case 2: Only 1 married couple is present in the committee.

We have 3 married couples and 1 couples to choose from, hence number of possibilities are: 3C1 => 3
Notice that both the bachelors would need to present and we can't have 3 married people together (simple proof would via the Pigeon-hole principle)
Since both bachelors need to present number of ways => 1
Married couples: We have 2 places to fill yet and 2 choices for each => 2*2 => 4
Total number of ways => 3*4 => 12

Case 3: No married couple present in the committee.

Again via pigeon hole principle this case isn't possible(Notice if we had <=5 seats in the committee then it would be possible).

Hence, total number of ways => 15 + 12 + 0 =>(D) 27

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3 married couples and 2 bachelors can be grouped in 8C6 different ways: 28.
There is only one not allowed combination when we get the 3 married couples.

28-1=27

Correct answer D

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16 Dec 2024, 18:30

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From a group consisting of 3 married couples and 2 bachelors
i.e. 3 x 2=6 married persons and 2 bachelor persons (Total 6+2= 8 people)

We have to form 6 member group and the restriction is "at most two married couples" ie. 4 persons

Thus, 8C6=(8x7)/2 =28 (E)

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Bunuel

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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We're asked at most two married couples,
So there are 3 cases:
No married couple: This is not possible since if we take 2 bachelors, we still require 4 people for a committee of 6, so at least one couple will be included since 3 married couples = 3*2=6, out of which we pick 4.

One couple: 2 bachelors can be selected in 1 way
Among the married couple, one couple can be selected in 1 way, from the remaining 2C1 * 2C1 * 3, so that we don't choose a couple, and *3 because any couple can be chosen among the 3 = 12 cases

Two couples: 1 out of 2 bachelors can be selected in 2C1 ways
Among married couples, 2 couples can be selected in 1 way, and of the remaining one person can be selected in 2C1 * 3 so ways, and *3 because one person from any couple can be chosen among the 3 = 6 cases
Now total cases bachelors * couples = 2*6 = 12 cases

Total cases = 12 + 12 = 24 cases

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Bunuel

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The trick is that you can split up married couples. Knowing that you can get your combos:
Couple 1 = C1 when they are split apart I will use C_11 and C_12 and I'll use B1 and B2 for the bachelors

C1 C2 B1 B2
C1 C3 B1 B2
C2 C3 B1 B2

Now in each of these 3 above I can replace B1 or B2 with the someone from the unused married couple ie:

C1 C2 B1 C_31
C1 C2 B1 C_32
C1 C2 C_31 B2
C1 C2 C_32 B2

That means I have 3 + (3*4) = 15 combos so far

Finally I can do the same to replace a couple with people from other couples:

C1 C_21 C_31 B1 B2
C1 C_21 C_32 B1 B2
C1 C_22 C_31 B1 B2
C1 C_22 C_32 B1 B2

That gives 12 more combos, 4 for each couple giving a final answer of 27

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First we should calculate all cases:

C ( 8, 6) = 8! / 6! * 2! = 28

Now the restrictions

1 coupes : 0, its not possible because we looking to fill 6 people and there are just 2 single

2 couples: 3, 3! / 2 ! = 3

Total

Total cases - Restrictions
28 - 25

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At most 2 married couples = all combinations of 6 out of 8 people - combinations with 3 married couples = 8C6 - 3C3 = 28-1 =27

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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 people = 3*2 + 2 = 8
Number of comittee formed if includes at most 2 married couple
= Total ways of choosing 6 people out of 8 for the comittee - number of ways where 3 couples selected forthe committees
= 8C6 - 1 (only 1 way to select a comittee of 6 among 6 married people) = 27

aviraj1703

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3 married couples and 2 bachelors,
a committee of 6 members is to be formed.

Total ways to form a committee = 8C6 = 8C2 = 28.

All 3 married couples are included = 3*(1/3) = 1.

Committees with at most 2 married couples included = 28 - 1 = 27.

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The answer is equivalent to (total combinations) - (total ways to choose 3 couples) = 8C6 - 1 = 27

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HELLO,
we have total 3 married couples and 2 bachelors ,so to form 6 member committee with most 2 married couples
from total ways i have reduced the unfavorable condition.
8C6-6C6= 27
hence answer is D

Hope it helps
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Bunuel

This question was provided by GMAT Club
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I think answer is D.

For at most two married couples there will be 3 conditions based on given constraints.
1st - No couple
2nd - 1 Couple
3rd - 2 Couple

lets assume there are 3 pairs AB, CD, EF and XX are 2 bachelors.

But no couple condition will not be possible as lets check extreme case.
X X A C D _ (at last place there will always be one pair)
So no couple condition fails.

now lets check for 2nd - 1 Couple.

Selecting 1 couple out of 3 is = 3C1

Therefore total outcomes for 1 couple = Selecting 1 couple out of 3 * Selecting 1 from remaining couple * Selecting 1 from another remaining couple * Selecting 2 bachelors

$= 3C1 * 2C1 * 2C1 * 2C2 = 3 * 2 * 2 * 1 = 12$

now lets check for 3rd - 2 Couple.

Total outcomes for 2 Couple = Selecting 2 out of 3 couple ((Selecting 1 out of Remaining 1 couple * Selecting 1 from remaining 2 bachelors) OR (Selection of Both Bachelors))

$= 3C2 * ( (2C1 * 2C1) + 2C2 ) = \frac{3*2 }{ 2!} * ( (2*2) + 1 ) = 3*5 = 15$

Total Cases = sum of all 3 conditions = 12 + 15 = 27

Lets Go Team Grinch!!!
Lets Win it!!!

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in atmost/ atleast questions one sense check is to see are we removing only a few cases... if you think yes... using total - (not A) = (A) helps!
using similar logic here... total cases= 8C6= 8x7/2=28 cases...
removing all 3 selected were married couples ...6C6=1
cases where at most 2 couples were selected are 28-1=27 (ans)

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There are 8 people in total.
We need to choose 6.
The number of ways to do this is 8C6 = 28.

Now there are 3 married couples, and we need to choose all 3.
This can be done in 3C3 = 1 way.

To get the number of committees with at most two married couples, we subtract the number of committees with three married couples from the total number of possible committees: 28 - 1 = 27

Therefore, the answer is D. 27

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