# There are 9 employees in a bank of whIch 6 are clerks, 2 are peons and 1 is a manager.

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There are 9 employees in a bank of whIch 6 are clerks, 2 are peons and 1 is a manager. In how many ways can a committee of 4 members be formed such that

(1 ) the manager must be selected?

(2 ) two peons are not to be selected and the manager Is to be selected?

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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