Imagine there are $5$ people around a round table: A, B, C, D and E. A and D must sit together. C and E must not sit together. How many different ways can they be seated?
I know that with A and D sitting next to each other there are $12$ different arrangements; $(5-1)! – ((4-1)! \times 2) = 12$. Due to treating A and D as one unit. How do I include C and E not sitting next to each other into the equation?
Thanks
Best Answer
We initially seat A, B, and D. Since A and B must sit together, there are only two ways to do this, depending on whether B sits to the left or right of A, as shown below.
To ensure that C and E do not sit together, we must place them to the left or right of the block. There are two ways to do this. Hence, there are four admissible arrangements, as shown below.