I have a particular type of combinatorics problem, that seems to be common, but i feel there is a bit more to this problem. Here is the question:
How many ways can 7 people be seated in a row of 9 chairs if Rob and Rhonda must sit next to each other for all possibilities.
I have seen people that have given answers to this problem, but almost everybody is getting a different answer. I saw one solution that did it in 3 portions:
1) Rob-Rhonda connected + Rhonda-Rod connected: = 8*2 = 16 ( I agree with this part)
2) Now removing Rob-Rhonda, we have 5 people left as well as 7 chairs left. Now they say: 7C5 = 21 (This one i am not sure about, because all people must be in a row, so I ASSUME that means there cannot be empty seats between people) This 7-Choose-5, I don't think will place people in a row per se.
3)Then the solution says when we place people in the first chair, this first chair can have a choice of 5 people, then the next chair can have 4, etc.
So the overall answer would be the multiplication of these parts:
(8*2) * 7C5 * 5! = 40 320.
Out of all the different solutions I have seen to this, I thought this might be the right one.
Hope someone here can explain this in more detail and/or tell me if this is correct answer or incorrect.
Here is another way to arrive at the same answer.
First, wrap Rob and Rhonda with duct tape and replace them with a single "object". Second, add two dummies to the group to fill the vacant chairs. We now have $8$ objects in all, which can be arranged in $8!$ ways. But we need to multiply by $2$ since the Rob-Rhonda pair can appear in either order, and we need to divide by $2$ because the two dummies are indistinguishable. So the final number of arrangements is $$\frac{8! \times 2}{2} = 8! = 40,320$$