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 RESTRICTIONSB. IF EACH COUPLE IS TO SEAT TOGETHERC. IF ALL THE MEN SEAT TOGETHER TO THE RIGHT ALL THE WOMEN
Hello, HANY!
WITH NO RESTRICTIONS, THERE ARE: .$\displaystyle 8! \,=\,40,\!320$ WAYS.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?
DUCT-TAPE THE COUPLES TOGETHER.B) If each couple is to sit together?
THEN WE HAVE FOUR "PEOPLE" TO ARRANGE: .$\displaystyle \boxed{Aa}\;\boxed{Bb}\;\boxed{Cc}\;\boxed{Dd}$
THERE ARE .$\displaystyle 4! \,=\,24$ POSSIBLE ARRANGEMENTS.
BUT EACH COUPLE COULD BE TAPED IN TWO WAYS: .$\displaystyle \boxed{Aa}\text{ or }\boxed{aA}$
. . HENCE, THERE ARE: .$\displaystyle 2^4 \,=\,16$ WAYS TO SEAT THE COUPLES.
THEREFORE, THERE ARE: .$\displaystyle 24\cdot16 \:=\:384$ WAYS.
THEY ARE SEATED LIKE THIS: .$\displaystyle W~W~W~W~M~M~M~M$C) If all the men sit together to the right of all the women?
THE FOUR MEN CAN BE SEATED IN: .$\displaystyle 4!\,=\,24\text{ WAYS.}$
THE FOUR WOMEN CAN BE SEATED IN: .$\displaystyle 4!\,=\,24\text{ WAYS.}$
THEREFORE, THERE ARE: .$\displaystyle 24\cdot24 \:=\:576\text{ ARRANGEMENTS.}$
Hello hany,
A. $\displaystyle 8!=40320$
B. $\displaystyle 4!*2^4= 384$ (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. $\displaystyle 4!*4!=576$ (4! ways in which women can take their seats $\displaystyle \times$ 4! in which men can take their seats)