A committee of 6 members is to be selected from a group of 8 men and 6 women in such as way that at least 3 men are there in the committee. In how many different ways can it be done ?

There are a total of 8 men and 6 women.

Number of ways in which the committee has exactly 3 men is $$^8C_3 \times ^6C_3 = 1120$$
Number of ways in which the committee has exactly 4 men is $$^8C_4 \times ^6C_2 = 1050$$
Number of ways in which the committee has exactly 5 men is $$^8C_5 \times ^6C_1 = 336$$
Number of ways in which the committee has exactly 6 men is $$^8C_6 \times ^6C_0 = 28$$

Hence, the total is $$1120 + 1050 + 336 + 28 = 2534$$