Question

In a small class, there are 10 students, 6 men and 4 women. Only four students can be accommodated on an exciting all expense paid field trip. Two must be men and two must be women. How many different sets of students could be selected for the trip? Please explain how to get the answer below using combination and / or permutation method(s).

Answer: 90

Answer 1

Answer 2

To solve this problem, we can use combinations.

We need to select two men out of six and two women out of four to form a group of four students.

The number of ways to select two men from six is denoted as C(6, 2) and can be calculated using the combination formula:

C(n, r) = n! / (r! * (n - r)!)

where n is the total number of items to choose from, and r is the number of items to be selected.

Similarly, the number of ways to select two women from four is denoted as C(4, 2).

To find the total number of different sets of students that could be selected for the trip, we need to multiply these two combinations together since the selections are independent:

Total number of sets = C(6, 2) * C(4, 2)

Calculating C(6, 2):

C(6, 2) = 6! / (2! * (6 - 2)!) = 6! / (2! * 4!) = (6 * 5 * 4!) / (2! * 4!) = (6 * 5) / (2 * 1) = 15

Calculating C(4, 2):

C(4, 2) = 4! / (2! * (4 - 2)!) = 4! / (2! * 2!) = (4 * 3 * 2!) / (2! * 2!) = (4 * 3) / (2 * 1) = 6

Calculating the total number of sets:

Total number of sets = C(6, 2) * C(4, 2) = 15 * 6 = 90

Therefore, there are 90 different sets of students that could be selected for the trip, where two students must be men and two students must be women.