combinations or permutations? 5 rows, each row capable of seating 2 people...

I have 5 rows, each row capable of seating 2 people, 1 left and 1 right and have 10 people to fill them with.
(Persons A, B, C, D, E, F, G, H, I, J).

However, 2 of which are always seated side by side, I and J, and their position within the 5 rows is denoted by rolling a d5 die.

When the 2 people are seated side by side in 1 of the rows, I can then place the remaining 8 people in the 8 remaining
seats at random.

Am I correct in suggesting that there are 8^8x5 possible seating arrangements?


Thank you in advance for your time and effort.

Kind regards,

James

James86 said:

I have 5 rows, each row capable of seating 2 people, 1 left and 1 right and have 10 people to fill them with. (Persons A, B, C, D, E, F, G, H, I, J).

However, 2 of which are always seated side by side, I and J, and their position within the 5 rows is denoted by rolling a d5 die.(??????) When the 2 people are seated side by side in 1 of the rows, I can then place the remaining 8 people in the 8 remaining seats at random.
Am I correct in suggesting that there are 8^8x5 possible seating arrangements?NO! ​

Click to expand...

The correct answer is \(\displaystyle 5\cdot 2\cdot (8!)\)
If you need more help, then you must reply with why you doubt that answer.

James86 said:

The correct answer is 5⋅2⋅(8!)
- I'm wondering if you could just write that in its simplest terms so I can at least get a sum from it.
Sorry about this.

Click to expand...

There are five ways to select a row for the two.
There are two ways to seat the two side-by-side.
There are (8!) ways arrange the remaining people.