1. ## easy, easy question

How many ways are there to seat 10 people, consisting of 5 couples, in a row of seats (10 seats wide) if all couples are to get adjacent seats?

2. Hello, billym!

How many ways are there to seat 10 people, consisting of 5 couples,
in a row of seats (10 seats wide) if all couples are to get adjacent seats?
For reference, let the couples be: .$\displaystyle (A,a),\:(B,b),\:(C,c),\:(D,d),\:(E,e)$

Duct-tape the couples together.

We have 5 "people" to arrange: .$\displaystyle \boxed{Aa}\;\boxed{Bb}\;\boxed{Cc}\;\boxed{Dd}\;\b oxed{Ee}$

There are: .$\displaystyle 5! \,=\,120$ permutations.

But for each permutation, the couples can be "swtiched".

. . $\displaystyle \boxed{Aa}$ could be $\displaystyle \boxed{aA}$, $\displaystyle \boxed{Bb}$ could be $\displaystyle \boxed{bB}$, and so on.

There are: .$\displaystyle 2^5 \,=\,32$ possible switchings.

Therefore, there are: .$\displaystyle 120\cdot32 \:=\:{\color{blue}3840}$ seating arrangements.

3. Originally Posted by Soroban
Hello, billym!

For reference, let the couples be: .$\displaystyle (A,a),\B,b),\C,c),\D,d),\E,e)$

Duct-tape the couples together.

I've always suspected you had an evil streak, Soroban!

I have a friend who swears you can do anything with duct-tape. Now I know you can even solve math problems with it!

We have 5 "people" to arrange: .$\displaystyle \boxed{Aa}\;\boxed{Bb}\;\boxed{Cc}\;\boxed{Dd}\;\b oxed{Ee}$

There are: .$\displaystyle 5! \,=\,120$ permutations.

But for each permutation, the couples can be "swtiched".

. . $\displaystyle \boxed{Aa}$ could be $\displaystyle \boxed{aA}$, $\displaystyle \boxed{Bb}$ could be $\displaystyle \boxed{bB}$, and so on.

There are: .$\displaystyle 2^5 \,=\,32$ possible switchings.

Therefore, there are: .$\displaystyle 120\cdot32 \:=\:{\color{blue}3840}$ seating arrangements.

4. Hello, HallsofIvy!

I have a friend who swears you can do anything with duct tape.
Duct tape is like The Force.

It has a light side and dark side
and it holds the universe together.