# Thread: Cross checking a simple permutation problem

1. ## Cross checking a simple permutation problem

Hi guys,

Q) In how many different ways can 6 gentlemen and 5 ladies sit around a table if no two ladies sit side by side?

My answer: (6-1)! * 5! = 5! * 5!

Am i correct?

2. Hello, saberteeth!

In how many different ways can 6 men and 5 women
sit around a table if no two ladies sit side by side?

Consider 12 chairs around the table.

Seat the six men in alternate chairs.
. . There are: . $(6-1)! \:=\:120$ ways.

Now seat the five women in five of the six empty chairs.
. . There are: . $_6P_5 \:=\:720$ ways.

Therefore, there are: . $120\cdot720 \:=\:86,\!400$ seating arrangements.

3. Originally Posted by Soroban
Hello, saberteeth!

Consider 12 chairs around the table.

Seat the six men in alternate chairs.
. . There are: . $(6-1)! \:=\:120$ ways.

Now seat the five women in five of the six empty chairs.
. . There are: . $_6P_5 \:=\:720$ ways.

Therefore, there are: . $120\cdot720 \:=\:86,\!400$ seating arrangements.

Hi Soroban,

Thanks for your reply. I am confused.. Wouldn't that be the same permutation for 6 men and "6 women"? I read it on the 2nd example on this website: All about Circular Permutations | TutorVista.com

4. Hello, saberteeth!

Wouldn't that be the same permutation for 6 men and 6 women?
Yes, it is!

Look at it this way . . .

There are 12 chairs.
Once we seat the 6 men and 5 women . . . there is one empty seat.

And that is where the 6th woman can sit.

A similar example:

There are six chairs in a row.
In how many ways can five men be seated?

The answer is a permutation: . $_6P_5 \:=\:720$ ways.

There are 6 chairs in a row.
In how many ways can six men be seated?

The answer is a permutation: . $_6P_6 \:=\:720$ ways.

Can you see why the answers are the same?