siddharthsinha123 wrote:
2 students from each of the 4 different B-schools participate in a debate competition. In how many different ways can 3 students be selected to represent at national level so that no 2 students are from the same school??
A. 26
B. 32
C. 35
D. 40
E. 50
We are given that there are 2 students from each of the 4 different B-schools (or 8 people) and we need to determine the number of ways to choose 3 students in which no 2 people are from the same school. So this is a special combination problem. Before we tackle this problem, let’s review a combination problem with no restrictions.
With no restrictions, the number of ways to choose 3 people from 8 is 8C3, which is calculated as follows:
8C3 = 8!/[3!(8-3)!] = (8 x 7 x 6)/3! = 56
In this (special combination) problem, 3 students are chosen in which no students from the same school can serve together. The first person could be any one of the 8 students. However, once a student is selected, the other student from the same school cannot be selected. This reduces the choice of the second student to 6 possible students. Once the second student is chosen, the other student from the same school cannot be selected. This reduces the number of students who could be chosen as the third student to 4. Therefore, the number of ways of choosing these 3 students is:
(8 x 6 x 4)/3! = 32
Answer: B