In how many ways can a committee of 5 be selected from 8 students if 2 particular students are to be included?
-
20
- 28
- 54
- 58
The correct answer is: A
Explanation
Since 2 students must be included, we have to arrange the remaining 3 students from the 6 students left
= \(^{6}C_{3} = \frac{6!}{(6-3)!3!}\)
= 20 ways