Question

a class consist of 12 boys of whom 5 are perfects. how many committee of 8 can be formed if its to have; a) 3 prefects b) at least 3 prefects​

Answer 1

Explanation:

a) To form a committee of 8 with 3 prefects, we need to choose 3 prefects from the 5 available prefects, and 5 non-prefects from the remaining 7 boys. We can do this by using the combination formula:

Number of ways = (Number of ways to choose 3 prefects) × (Number of ways to choose 5 non-prefects)

Number of ways to choose 3 prefects from 5 = C(5, 3) = 10

Number of ways to choose 5 non-prefects from 7 = C(7, 5) = 21

Therefore, the total number of committees of 8 with 3 prefects that can be formed is:

Number of ways = 10 × 21 = 210

b) To form a committee of 8 with at least 3 prefects, we need to consider two cases: one where we choose exactly 3 prefects, and one where we choose all 5 prefects. We can calculate the number of ways for each case using the combination formula:

Number of ways to choose exactly 3 prefects and 5 non-prefects = C(5, 3) × C(7, 5) = 210

Number of ways to choose all 5 prefects and 3 non-prefects = C(5, 5) × C(7, 3) = 35

Therefore, the total number of committees of 8 with at least 3 prefects that can be formed is:

Number of ways = (Number of ways to choose exactly 3 prefects and 5 non-prefects) + (Number of ways to choose all 5 prefects and 3 non-prefects)

Number of ways = 210 + 35 = 245