Problem Solving

willbeatthegmat wrote:In how many ways can 5 girls & 3 boys be seated in a row such that no 2 boys are together?

Here's how I would set this up:

Have 11 chairs and first seat the girls in seats 2, 4, 6, 8, and 10
_G_G_G_G_G_
This can be accomplished 5! ways.

Note: This arrangement prevents the boys from sitting together.
Now seat each of the 3 boys in one of the 6 remaining seats (then remove the 3 empty seats).
This can be accomplished in 6x5x4 ways

So, the total number of ways to seat all of the boys and girls is 5! x 6x5x4

Case II
Now, we need to find 2 boys sitting together (which is what originally is asked)
Here again treat the 2 boys as 1 seat,
7 * 6 * 5 * 4 * 3 * 2 * 1 = 7! ways
But again there are 2 ways the chosen boys can arrange themselves and another 3 ways the 2 boys can be chosen
So the total number of ways = 7! * 2 * 3 = 7! * 6
[we need to subtract the 12 ways in which all 3 would sit together, which is already included above in Case I]
The total number of ways = (7! * 6) -12 ways

This approach will work; however, there are a few problems with your calculations:

In case II you counted each BBB situation twice.
To see how this happened, we'll let the 3 boys be A, B, and C. If we follow your solution and consider the case where A and B are combined into one child (AB), we could have one possible seating arrangement of GGG(AB)CGG. In another situation (which you have counted separately) we can combine B and C together to form one child (BC). In this situation, we can have the seating arrangement GGGA(BC)GG

As you can see, we have counted the arrangement GGGABCGG twice. So, from your total of 7!*6, we need to subtract all of the BBB arrangements since these situations where all 3 boys are seated together have been counted twice. How many BBB situations are there? There are 6!*3! (as calculated in case I)

So, the final value for case II should have been 7!*6 - 6!*3!

Finally, we need to recognize that all of the exceptions (at least 2 boys together) are already accounted for in case II. We don't need to include case I (although we did need it to determine the total value for case II)

So, the final answer is 8! - (7!*6 - 6!*3!)

You will find that this answer is the same as my answer of 5! x 6x5x4