(i) In this case we have to select one girl out of 6 and 3 boys out of 5.

The number of ways of selecting 3 boys = ^{5}C_{3} = ^{5}C_{2} = 10.

The number of ways of selecting one girl = ^{6}C_{1} = 6.

∴ The required committee can be formed in 6 × 10 = 60 ways.

(ii) The committee can be formed with (a) one boy and three girls, or (b) 2 boys and 2 girls,

or (c) 3 boys and one girl, or (d) 4 girls alone.

The committee can be formed in (a) ^{5}C_{1} × ^{6}C_{3} ways ;

The committee can be formed in (b) ^{5}C_{2} × ^{6}C_{2} ways ;

The committee can be formed in (c) ^{5}C_{3} × ^{6}C_{1} ways ;

The committee can be formed in (d) ^{6}C_{4} ways.

Hence the required number of ways of forming the committee

= ^{5}C_{1} × ^{6}C_{3} + ^{5}C_{2} × ^{6}C_{2} + ^{5}C_{3} × ^{6}C_{1} + ^{6}C_{4} = 100 + 150 + 60 + 15 = **325** **ways. **

**(ii) Method II.** Required ways = (Committees of 4 out of 11 without any restriction) – (Committees in which no girl is included) = ^{11}C_{4} – ^{5}C_{4} =** 325.**