I am having trouble with this maths problem:
There are 8 people at a round table - 6 guests, 2 hosts. How many possible arrangements are there if the 2 hosts aren't allowed to sit next to each other.
Thanks
hint:
sit the two host first, then let the guest find their seats. So, there are 8 choices for the 1st host, 5 remaining choices for the 2nd host, 6 remaining choices for the 1st guest, 5 for the 2nd guest, and so on.
Take it from here.
To solve this problem, we can follow the provided hint.
First, we will need to find the number of ways to arrange the two hosts without any restrictions. Since they are distinct (one is the 1st host and the other is the 2nd host), there are 8 choices for the 1st host and 7 remaining choices for the 2nd host. Therefore, there are 8 * 7 = 56 possible arrangements for the hosts.
Next, we can consider the arrangement of the remaining guests. After the two hosts have been seated, there are 6 guests left to be seated. Since the table is round, there will always be a guest sitting to the left and right of each person.
Now, to calculate the number of ways to arrange the remaining guests, we can start with the first guest. There are 6 seats available for the first guest to sit. After the first guest has been seated, there will be 5 remaining seats for the second guest. For each successive guest, the number of available seats will decrease by one.
Therefore, the total number of arrangements for the guests is 6 * 5 * 4 * 3 * 2 * 1 = 720.
Finally, to find the total number of arrangements where the hosts are not allowed to sit next to each other, we can multiply the number of arrangements for the hosts with the number of arrangements for the guests.
Total number of arrangements = 56 * 720 = 40,320 possible arrangements.
So, there are 40,320 possible seating arrangements if the two hosts are not allowed to sit next to each other at the round table.