Rectangular Table Arrangement

a) In how many ways can 13 people be seated on one side of a rectangular table if Doug refuses to sit next to Gordon?

I have two different ideas-

Idea 1) There are two options: either Doug is at the end of the table or Doug is in the middle of the table. If Doug is at the end: He has 2 seats to choose from and Gordon would have 11 seats to choose from so $2 \times 11$. If Doug is in the middle of the table somewhere: Doug has 11 seats to choose from and then Gordon can sit any 10 of the seats. So $11\times10$. So in total we would have $2 \times 11+11\times10$. I know this is not right and I'm counting the combinations wrong, but I think the idea could work.

Idea 2) My second idea is to start with $13!$ and then subtract the number of ways Doug and Gordon are sitting next to each other.

b) What if Doug insists on sitting to the right of Gordon (not necessarily next to him)?

First, there are 12 seat options for Doug to sit in (It can't be 13 because if Doug sits in the rightmost chair then Gordon can't sit to the right of Doug) and then there are $12!$ ways to place Gordon to the right of Doug. So the answer would be $12\times 12!$.Does this seem to make sense?

In how many ways can $13$ people be seated on one side of a rectangular table if Doug sits next to Gordon?

We count the number of ways of arranging $13$ people in a row, then exclude those arrangements in which Doug sits next to Gordon.

In how many ways can $13$ people be arranged in a row?

$$13!$$

To calculate the number of arrangements in which Doug and Gordon sit next to each other, place Doug and Gordon in a box. Then you have $12$ objects to arrange, the box containing Doug and Gordon and the other $11$ eleven people. In how many ways can the $12$ objects be arranged?

$$12!$$

In how many ways can Doug and Gordon be arranged within the box?

$$2!$$

In how many arrangements do Doug and Gordon sit next to each other?

$$12!2!$$

In how many arrangements of the $13$ people do Doug and Gordon not sit next to each other?

$$13! - 12!2!$$

As for the second problem:

In how many ways can $13$ people be seated on one side of a rectangular table if Doug insists on sitting to the right of Gordon?

By symmetry, Doug sits to the right of Gordon in exactly half of the possible arrangements.

$$\frac{1}{2} \cdot 13!$$

• Interesting -- to my mind, bringing randomness into the symmetry argument is an unnecessary contamination :-) Commented Feb 26, 2016 at 22:14

In your first idea, you're only counting configurations of Doug and Gordon and not of the other $11$ people.

Your second idea is how I'd approach the problem – keeping in mind that again "the number of ways Doug and Gordon are sitting next to each other" must take into account everyone and not just Doug and Gordon.

For b) your answer $12\cdot12!$ is wrong. By symmetry, exactly half of all configurations have Doug sitting to the right of Gordon, so the number of seatings in this case is $\frac12\cdot13!$.

• So could I do $13!-24$? 24 being the number of ways Gordon and Doug are right next to each other. How do I need to take everyone else into account? Commented Feb 26, 2016 at 22:06
• @user2553807: It seems you haven't understood yet what was wrong with your calculation for idea 1). You can't just count the number of ways Doug and Gordon can sit -- for each of those ways, the remaining $11$ people can sit in $11!$ different ways. Your count of $24$ for Doug and Gordon is correct -- taking into account the other people, the result is $13!-24\cdot11!=13\cdot12!-2\cdot12!=11\cdot12!$. Commented Feb 26, 2016 at 22:08