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

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

Scenario 1: The couple is on 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 ways to select the group is 28 + 70 = 98 ways.

Answer: A

Yes it is, sometimes i get very stuck with these questions. Thanks!

Hi All,

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:

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