# Math Help - permutation and combinations

1. ## permutation and combinations

3 men and 4 women are seated at a round table. Find the number of possible arrangements if one particular woman must sit between 2 men.

2. ## Re: permutation and combinations

Originally Posted by Punch
3 men and 4 women are seated at a round table. Find the number of possible arrangements if one particular woman must sit between 2 men.

Since table is round answer is :

$6\cdot 4!$

3. ## Re: permutation and combinations

Originally Posted by princeps
Since table is round answer is :

$6\cdot 4!$
Yes, the answer is right but I dont understand why it is 6x4

4. ## Re: permutation and combinations

In addition, what is wrong with 4!2!? I don't see how it is wrong...

5. ## Re: permutation and combinations

Originally Posted by Punch
Yes, the answer is right but I dont understand why it is 6x4
Actually correct formula is :

$\frac{3!}{(3-2)!}\cdot 4!$

6. ## Re: permutation and combinations

Originally Posted by princeps
Actually correct formula is :

$\frac{3!}{(3-2)!}\cdot 4!$
I see, I understand the 4! part, but I still dont get the 3!/(3-2)! part.

7. ## Re: permutation and combinations

Originally Posted by Punch
I see, I understand the 4! part, but I still dont get the 3!/(3-2)! part.
That part represents number of permutations with no repetition .

You have 3 men and 2 seats . Do you understand now ?

8. ## Re: permutation and combinations

Originally Posted by princeps
That part represents number of permutations with no repetition .

You have 3 men and 2 seats . Do you understand now ?
So there are 3 men vying for 2 seats, but I still dont understand how it is worked out

9. ## Re: permutation and combinations

Originally Posted by Punch
So there are 3 men vying for 2 seats, but I still dont understand how it is worked out
There are three ways to select the two men who will sit on either side of that woman. There are two ways the seat those three. That gives six. Put that group at the table. Now there are 4! ways to seat the other four.

10. ## Re: permutation and combinations

Originally Posted by princeps
That part represents number of permutations with no repetition .

You have 3 men and 2 seats . Do you understand now ?
So that is 3P2 whereby 2 man are selected to sit beside the woman. Why is there no 4C1 included to select 1 woman from the 4 to sit in between the men?

Which gives, (choose 2men from 3)*(permute the 2men)*(choose 1 woman from 4)*(Permute in a circle)=3C2*2!*4C1*4!=576

11. ## Re: permutation and combinations

Hello, Punch!

3 men and 4 women are seated at a round table.
Find the number of possible arrangements
if one particular woman must sit between 2 men.

Since we have a round table, that woman can sit anywhere.

She will be flanked by two men.
There are 3 choices for the man on her right, 2 choices for the man on her left.

The remaining 4 people (one man, 3 women) can be seated in $4!$ ways.

Therefore, there are: . $3\cdot 2\cdot 4! \:=\:144$ arrangements.