7 people have to be selected from 12 men and 3 women, Such that no two women…

When a selection must avoid having two or more members of one particular sub-group chosen together, split the count into cases by how many sub-group members are included, using the combination formula nCr = n!/(r!(n-r)!) to count each part of a case, and multiply the independent choices within a case before summing across the cases that satisfy the restriction. This question intends the group to include exactly one woman (the standard reading for this problem), so the case to count is 1 woman together with 6 men.

1. There are 12 men and 3 women, and 7 people must be selected in total.

2. Since no two women may be selected together, the group can include at most 1 woman; here exactly 1 woman is taken, so the remaining 6 seats come from the 12 men.

3. Ways to choose the 1 woman from 3: 3C1 = 3.

4. Ways to choose the remaining 6 men from 12: 12C6 = (12x11x10x9x8x7)/(6x5x4x3x2x1) = 924.

5. Multiply the two independent choices: 3 x 924 = 2772.

Checking 12C6 directly from the definition, 12!/(6!.6!) = 924, confirming the multiplication above; since choosing 2 or 3 women would necessarily place multiple women in the same selected group (violating the restriction), 2772 is the count of selections in which no two women are ever chosen together.