There are 12 students to be seated in a row of 12 chairs. Paul and Martha have broken up and will not sit together. How many different seating arrangements are possible?

Were are the answers?

12!-(11!x2) = 399168000 is what i got

To solve this problem, we can use the concept of permutation and subtraction principle.

First, let's consider the total number of seating arrangements if Paul and Martha were not restricted to sit separately. Since there are 12 students and 12 chairs, there are 12! (12 factorial) ways to arrange them without any restrictions.

Now, let's find the number of seating arrangements where Paul and Martha sit together. To do this, we can treat Paul and Martha as a single entity. So, we can arrange Paul and Martha among themselves in 2! (2 factorial) ways, and then arrange the remaining 10 students in the remaining 10 chairs in 10! ways. Hence, the number of seating arrangements where Paul and Martha sit together is 2! * 10! = 2 * 10! (2 times 10 factorial).

However, we need to subtract this count from the total number of seating arrangements to get the final answer since we want to find the number of seating arrangements where Paul and Martha do not sit together.

So, the number of seating arrangements where Paul and Martha do not sit together is given by 12! - 2 * 10!.

Let's calculate this:

12! = 479,001,600
10! = 3,628,800

Now, let's substitute these values into the equation:

12! - 2 * 10! = 479,001,600 - (2 * 3,628,800)
= 479,001,600 - 7,257,600
= 471,744,000

Therefore, there are 471,744,000 different seating arrangements possible where Paul and Martha do not sit together.