# easy, easy question

• Jul 14th 2009, 07:06 AM
billym
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?
• Jul 14th 2009, 07:20 AM
Soroban
Hello, billym!

Quote:

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: . $(A,a),\:(B,b),\:(C,c),\:(D,d),\:(E,e)$

Duct-tape the couples together.

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

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

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

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

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

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

• Jul 14th 2009, 07:55 AM
HallsofIvy
Quote:

Originally Posted by Soroban
Hello, billym!

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

Duct-tape the couples together.

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

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

Quote:

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

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

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

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

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

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

• Jul 14th 2009, 08:38 AM
Soroban
Hello, HallsofIvy!

Quote:

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.