It is given in the question that from a group of 8 men and 7 women, a committee of 6 is chosen.
And we have to find the number of ways can the committee be chosen so that the same amount of men and women are chosen. That is number of men= number of women =3
Selecting 3 men out of 8 men is 8C3 which is equal to 56.
Selecting 3 women out of 6 women is 6C3 which is equal to 20 .
So the required selection is the product of 56 and 20 that is 1120.