Out of 9 employces In a bank. 6 are clerks. 2 are peons and 1 Is a manager. A committee of 4 members Is to be formed.
(1) The manager must be selected In the committee:
If the manager Is to be selected In the committee of 4 members, than from the remaining (6 clerks + 2 peons) 8 employees, the remaining 3 members of the committee can be selected in 8C3 ways.
∴ Total combinations = 1C1 × 8C3
= 1 × \(\frac{8×7×6}{3×2×1}\)
= 56
(2) Two peons are not be selected and the manager is to be selected:
If the manager Is to be selected In the committee of 4 mcmbcrs and two peons are not be selected than from the remaIning 6 clerks, the remaining 3 members of the committee can be selected in6C3 ways.
∴ Total combinations = 1C1 × 6C3
= 1 × \(\frac{6×5×4}{3×2×1}\)
=20