In a class of 10 students, a group of 4 will be selected for

[BTGmoderatorAT]

In a class of 10 students, a group of 4 will be selected for a trip. How many different groups are possible, if 2 of those 10 students are a married couple and will only travel together?

A. 98
B. 115
C. 122
D. 126
E. 165

I'm confused between A and C. Can any experts help?

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Hi ardz24,

We're told in a class of 10 students, a group of 4 will be selected for a trip. We're asked for the number of different groups that are possible, if 2 of those 10 students are a married couple and will only travel together. Since we're dealing with groups, we'll use the Combination Formula (a couple of times) to answer this question.

Given the 'restriction' in the prompt (about the married couple), there are 2 types of groups to consider:
1) Groups WITH the married couple
2) Groups WITHOUT the married couple

WITH the married couple, there will be 2 'spots' for the remaining 8 people, so there are 8c2 = 8!/6!2! = 28 different groups
WITHOUT the married couple, there will be 4 'spots' for the remaining 8 people, so there are 8c4 = 8!/4!4! = 70 different groups

Total possible groups = 28+70 = 98

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

[Scott@TargetTestPrep]

In a class of 10 students, a group of 4 will be selected for a trip. How many different groups are possible, if 2 of those 10 students are a married couple and will only travel together?

A. 98
B. 115
C. 122
D. 126
E. 165

I'm confused between A and C. Can any experts help?

There are two scenarios, one in which the married couple is selected for the trip and the other in which it is not.

Scenario 1: The couple is selected for the trip

Since both the husband and the wife must be together, that leaves 8 students for 2 places, which can be determined in 8C2 = 8!/[2!(8-2!] = 8!/(2!6!) = (8 x 7)/2! = 28 ways.

Scenario 2: The couple is not on the trip

Since the married couple is not considered, that leaves 8 students for 4 places, which can be determined in 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (8 x 7 x 6 x 5)/(4 x 3 x 2) = 7 x 2 x 5 = 70 ways.

Thus, the total number of ways to select the group is 28 + 70 = 98 ways.

Answer: A

[swerve]

In a class of 10 students, a group of 4 will be selected for a trip. How many different groups are possible, if 2 of those 10 students are a married couple and will only travel together?

A. 98
B. 115
C. 122
D. 126
E. 165

I'm confused between A and C. Can any experts help?

We need to break the grouping down to 2 scenarios: with the married couple or without the married couple in the group of 4 people for the trip.

1) With the married couple in the group, we have to find the number of possibilities for the other 2 spots in the group from the left 8 people \(= \frac{8!}{2!\cdot 6!} = 28\)
2) Without the married couple in the group, we have to find the number of possibilities for picking 4 people from the 8 unmarried people from the group \(= \frac{8!}{4!\cdot 4!} = 70\)

Total number of possibilities = 28 + 70 = 98\(\,\Rightarrow\) A

[Aditi Goyal]

Since the couple will always and only travel together

Two types of groups can be made :
1) Group with the married couple
2) Group without the married couple

With the married couple, there will be 2 'spots' for the remaining 8 people, so there are 8c2 = 8!/6!2! = 28 different groups
Without the married couple, there will be 4 'spots' for the remaining 8 people, so there are 8c4 = 8!/4!4! = 70 different groups

Total possible groups = 28+70 = 98

Correct option is A