# (GR. 10) 10 people are to be seated in a row. What is the total number of ways if…

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.

• Please show us your calculation. – saulspatz Feb 23 at 14:54
• I think it should be $8!*8*2$.I think your answer is correct. Cheers :) – Abhinav Feb 23 at 15:00

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.

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.

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.

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
$$8!\cdot 8\cdot 2 = 645120.$$

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?