Nusa84 wrote:

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

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

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