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Please help me! I understand what the question is asking for, but I can’t seem to get the right answer. The correct no. of ways should be $645,120$, though that may be incorrect. If anyone is kind enough to show me the solution, I would be very grateful.

$10$ people are to be seated in a row. What is the total number of ways in which this can be done if Eric and Carlos always have exactly one of the other people sitting between them?”

EDIT: Oh wow that was fast! Thank you for your kind hints! I was finally able to get the answer.

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    $\begingroup$ Please show us your calculation. $\endgroup$ – saulspatz Feb 23 at 14:54
  • $\begingroup$ I think it should be $8!*8*2$.I think your answer is correct. Cheers :) $\endgroup$ – Abhinav Feb 23 at 15:00
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The possible positions of the two people are $$1-3,2-4,\cdots ,8-10$$ that is $8$ possibilities. We can swap the places, so multiply with $2$. Then, multiply with $8!$ because the other people can have $8!$ possible orders.

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Here's a hint to get started. Suppose Alice is seated between Eric and Carlos. Then we can treat Eric-Alice-Carlos as a block to be arranged with the other $7$ students.

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There are a total of $10$ people so there are $8$ people who could be seated between Eric and Carlos. There are $2$ ways of seating "Eric, other person, Carlos" or "Carlos, other person, Eric". Now treat those $3$ people as a single "person"- there are $8!$ ways to seat those $8$ "people". There are, then, $$ 8!(2)(8)= 645120 $$ ways to do this. That is the same as Peter's answer.

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Well Eric and Carlos always sits with one person between them. So let us assume Eric, Carlos and the third person as a single person. So we have $8$ people to be arranged. This can be done in $8!$ ways. Further there can be $8$ different people between Eric and Carlos. Also Eric and Carlos can be arranged between them in two different ways. So the answer is
$$8!\cdot 8\cdot 2 = 645120.$$

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HINTS:

Let us denote Eric as $E$ and Carlos as $C$.

We always need to have at least one other person in between these and this person can be chosen in $8$ different ways.

Glue together Eric, the chosen person (will be denoted as $X$) and Carlos into one block $EXC$. Since they could also sit as $CXE$ we are going to multiply the result we get by a factor of $2$.

Now you have $7$ people and the block of our glued together people.

Can you continue?

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