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?
out of the 12! possible seatings, there are 2*11! which seat them together. (Think of the two as a single unit.)
To find the number of different seating arrangements, we need to consider the different scenarios where Paul and Martha are either sitting next to each other or not sitting next to each other.
Let's first calculate the total number of seating arrangements without any restrictions. Since there are 12 students, we have 12 choices for the first seat, then 11 for the second seat, then 10 for the third, and so on until the last seat. This can be represented as 12!.
Now let's consider the scenario where Paul and Martha are sitting next to each other. We can treat them as one unit, so we have 11 units to arrange (11 students plus the Paul-Martha unit). Within this unit, we can arrange Paul and Martha in 2 different ways (Paul first, Martha second or Martha first, Paul second), and the remaining 11 students in 11! ways. This gives us 2 * 11!
Next, let's consider the scenario where Paul and Martha are not sitting next to each other. In this case, we break down the problem into two subproblems:
- First, we arrange the remaining 10 students (excluding Paul and Martha) in the 10 seats. We have 10! ways to do this.
- Then, we place Paul and Martha in any two of these 11 positions (the 10 seats plus the two ends). So, we have 11 available positions to place Paul and Martha.
This gives us 10! * 11 arrangements.
Finally, we can find the total number of seating arrangements by subtracting the unfavorable arrangements (where Paul and Martha sit next to each other) from the total number of arrangements:
Total arrangements - Unfavorable arrangements = 12! - (2 * 11! + 10! * 11)
Calculating this expression gives us the desired number of different seating arrangements.