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 ^{8}C_{3} ways.

∴ Total combinations = ^{1}C_{1} × ^{8}C_{3}

= 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 in^{6}C_{3} ways.

∴ Total combinations = ^{1}C_{1} × ^{6}C_{3}

= 1 × \(\frac{6×5×4}{3×2×1}\)

=20