In how many ways can 5 men and 7 women be seated in a row so that no two men are next to each other? You must justify your answer.

Answer 1

There are different ways to approach this problem, but one possible method is:

First, we can fix the positions of the women and consider the men as “dividers” between them. There are 8 gaps between the 7 women (including the ends), and we want to place the 5 men in such a way that no two of them are in adjacent gaps. To do that, we can use the principle of inclusion-exclusion: start with all possible arrangements (which is the number of ways to arrange 5 men and 7 women in a row, without any restriction), then subtract the arrangements where at least one pair of men are adjacent (which we can count by treating such a pair as a “block” and arranging the resulting 4 blocks and 7 women), then add the arrangements where at least two pairs of men are adjacent (which we can count by arranging the 3 “blocks” containing pairs of adjacent men and the 7 women), and so on, up to adding or subtracting the arrangements where all 5 men are adjacent (which is just the number of ways to arrange 6 objects, with the group of 5 men considered as one object). This gives:

Total arrangements = (12 choose 5) = 7920

Subtract arrangements with at least one adjacent pair of men = (7 choose 1) x (8 choose 4) x (5!) = 16800

Add arrangements with at least two adjacent pairs of men = (7 choose 2) x (7 choose 1) x (6!) = 15120

Subtract arrangements with at least three adjacent pairs of men = (7 choose 3) x (6 choose 2) x (5!) = 1680

Add arrangements with at least four adjacent pairs of men = (7 choose 4) x (5 choose 3) x (4!) = 280

Subtract arrangements with all five men adjacent = (7 choose 5) x (4!) = 168

Thus, the number of ways to seat the 5 men and 7 women in a row with no two men adjacent is:

7920 - 16800 + 15120 - 1680 + 280 - 168 = 1776

Therefore, there are 1776 possible arrangements.