3
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ I would say A and D sitting together has $2\times 3!=12$ ways where the $2$ is whether D is to A's left or right and the $3!$ is the ways of ordering B, C, E in the other three seats. But how many of these have C and E separate? $\endgroup$
    – Henry
    May 25, 2019 at 19:45
  • $\begingroup$ It could be C B E or E B C i think $\endgroup$
    – Randomer
    May 25, 2019 at 19:49
  • $\begingroup$ and so the answer is simply $2 \times 2=4$ $\endgroup$
    – Henry
    May 25, 2019 at 19:56

2 Answers 2

1
$\begingroup$

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.

initial_arrangements

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.

final_arrangements

$\endgroup$
1
$\begingroup$

First let's consider all permutations with $A$ and $B$ sitting together and then we'll subtract the ones where $C$ and $E$ are also sitting together.

$$\text{No. of permutations when A and B sit together} = (4-1)! 2! = 12$$

Now the ones where $C$ and $E$ are also sitting together are: $$\text{No. of permutations when A and B sit together AND C and E sit together} = (3-1)!2!2! = 8$$

Therefore the required no. of permutations are:

$$12-8 = 4$$

$\endgroup$
2
  • 1
    $\begingroup$ Thank you this has helped a lot! $\endgroup$
    – Randomer
    May 25, 2019 at 19:50
  • $\begingroup$ Sure. You're welcome. $\endgroup$
    – Vizag
    May 25, 2019 at 19:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .