# Math Help - Permutation/combination?

1. ## Permutation/combination?

1. FOUR MARRIED COUPLES HAVE BOTH 8 SEATS IN A ROW FOR A CONCERT. IN HOW MANY DIFF WAYS THEY CAN BE SEATED

A. WITH NO RESTRICTIONS
B. IF EACH COUPLE IS TO SEAT TOGETHER
C. IF ALL THE MEN SEAT TOGETHER TO THE RIGHT ALL THE WOMEN

2. Hello, HANY!

Four married couples have 8 seats in a row for a concert.
In how many different ways can they be deated:

A) with no restrictions?
WITH NO RESTRICTIONS, THERE ARE: . $8! \,=\,40,\!320$ WAYS.

B) If each couple is to sit together?
DUCT-TAPE THE COUPLES TOGETHER.
THEN WE HAVE FOUR "PEOPLE" TO ARRANGE: . $\boxed{Aa}\;\boxed{Bb}\;\boxed{Cc}\;\boxed{Dd}$
THERE ARE . $4! \,=\,24$ POSSIBLE ARRANGEMENTS.

BUT EACH COUPLE COULD BE TAPED IN TWO WAYS: . $\boxed{Aa}\text{ or }\boxed{aA}$
. . HENCE, THERE ARE: . $2^4 \,=\,16$ WAYS TO SEAT THE COUPLES.

THEREFORE, THERE ARE: . $24\cdot16 \:=\:384$ WAYS.

C) If all the men sit together to the right of all the women?
THEY ARE SEATED LIKE THIS: . $W~W~W~W~M~M~M~M$

THE FOUR MEN CAN BE SEATED IN: . $4!\,=\,24\text{ WAYS.}$
THE FOUR WOMEN CAN BE SEATED IN: . $4!\,=\,24\text{ WAYS.}$

THEREFORE, THERE ARE: . $24\cdot24 \:=\:576\text{ ARRANGEMENTS.}$

3. Originally Posted by hany

1. FOUR MARRIED COUPLES HAVE BOTH 8 SEATS IN A ROW FOR A CONCERT. IN HOW MANY DIFF WAYS THEY CAN BE SEATED

A. WITH NO RESTRICTIONS
B. IF EACH COUPLE IS TO SEAT TOGETHER
C. IF ALL THE MEN SEAT TOGETHER TO THE RIGHT ALL THE WOMEN
Hello hany,

A. $8!=40320$
B. $4!*2^4= 384$ (4! ways in which couples can be seated. M,F or F,M so there are two ways in which each couple can take their seats)
C. $4!*4!=576$ (4! ways in which women can take their seats $\times$ 4! in which men can take their seats)