In how many ways can a committee of 5 be selected from 8 students if 2 particular students are to be included?
-
A.
20 -
B.
28 -
C.
54 -
D.
58
Correct Answer: Option 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