Question

9. In how many ways can six men and six women
be seated in a row in
(a) any person may sit next to any other?
{b) men and women must occupy alternate
seats?​

Answer 1

Answer:

(a)12!

(b)2(6!)^2

Step-by-step explanation:

(a)

well there are 12 seats so

the first seat has 12 options

the second 11

the third 10

...

the last 1

so we have

$ways=12*11*10*\cdots *1=12!$

(b)

we can make two cases

case 1:

when the first one is a man

the first seat has 6 options (6 men)

the second seat has 6 options (6 women)

the third seat has 5 options (5 men)

...

the last seat has 1 option (1 woman)

$case_1 = 6*6*5*5\cdots 1=(6!)^2$

case 2

when the first one is a woman

it's the same analogy

so we have

$(6!)^2$

add both cases and get

$2(6!)^2$