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.

My answer: (5-1)!2! but answer is 144

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- Dec 1st 2011, 01:12 AMPunchpermutation 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.

My answer: (5-1)!2! but answer is 144 - Dec 1st 2011, 01:31 AMprincepsRe: permutation and combinations
- Dec 1st 2011, 01:40 AMPunchRe: permutation and combinations
- Dec 1st 2011, 01:43 AMPunchRe: permutation and combinations
In addition, what is wrong with 4!2!? I don't see how it is wrong...

- Dec 1st 2011, 02:00 AMprincepsRe: permutation and combinations
- Dec 1st 2011, 02:40 AMPunchRe: permutation and combinations
- Dec 1st 2011, 02:47 AMprincepsRe: permutation and combinations
- Dec 1st 2011, 02:54 AMPunchRe: permutation and combinations
- Dec 1st 2011, 03:14 AMPlatoRe: permutation and combinations
- Dec 1st 2011, 05:32 AMPunchRe: permutation and combinations
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 - Dec 1st 2011, 07:22 AMSorobanRe: permutation and combinations
Hello, Punch!

Quote:

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 $\displaystyle 4!$ ways.

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