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.
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.
To solve this problem, we can use the concept of permutations.
Step 1: Calculate the number of ways to arrange the women.
Since there are 7 women, they can be arranged in 7! (7 factorial) ways.
Step 2: Place the men between the women.
There are 8 possible positions between the women to place the 5 men. These positions are shown:
_ W _ W _ W _ W _ W _ W _
The first and last positions are used as options for the men to be seated. Using the concept of combinations, we can select 5 of these 8 positions to place the men. The number of combinations is given by 8 choose 5, written as C(8,5) or 8C5.
Step 3: Calculate the number of ways to arrange the men.
Since there are 5 men, they can be arranged in 5! (5 factorial) ways.
Step 4: Multiply the results from steps 1, 2, and 3 to get the total number of ways.
The total number of ways to arrange the men and women is given by: 7! x 8C5 x 5!
Calculating these values, we have:
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040
8C5 = 8! / (5! x 3!) = (8 x 7 x 6) / (3 x 2 x 1) = 56
5! = 5 x 4 x 3 x 2 x 1 = 120
So, the total number of ways to seat the 5 men and 7 women in a row such that no two men are next to each other is: 5,040 x 56 x 120 = 16,800,000