# another probability: number of ways for 4 girls and 4 boys to seat in a row

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Ej-lp ACayabyab
another probability: number of ways for 4 girls and 4 boys to seat in a row

In how many ways can 4 girls and 4 boys be seated in a row containing 8 seats if boys and girls must sit in alternate seats?
The answer in my notes is 1,152..
I need a solution please.

KMST

Let say the seats are numbered 1 through 8.
A boy could sit on seat number 1, and then all other boys would be in odd number seats,
or all the girls would sit in the odd numbered seats.
That is 2 possible choices.

For the people who will sit in the odd number seats,
you have to choose which of the 4 sits in seat number 1, who of the remaining 3 sits in seat number 3, and who of the remaining 2 sits in seat number 5. No more choices there, because the last one goes in seat number 7.
So there are 4*3*2=24 ways to arrange the people sitting in the odd number seats.

There are also 24 ways to arrange the people sitting in the even number seats (seats number 2, 4, 6, and 8).

With 2 ways to decide if seat number 1 is for a boy or a girls,
24 ways to arrange the boys,
and 24 ways to arrange the girls, there are
2*24*24=1152 possible seating arrangements.

Ej-lp ACayabyab

thank you again sir..clarify ko lang po sir:
san po nakuha ang 2 sa operation na ito: 2*24*24, wherein ung 24 each for boys and girls?

Jhun Vert

As an alternate solution, you can also think this problem as two benches, each can accommodate 4 persons. Say bench A and bench B. If boys will sit on A, girls are on B and conversely.

First Case: Boys at A, Girls at B
Number of ways for boys to seat on bench A = 4!
Number of ways for girls to seat on bench B = 4!
First Case = (4!)(4!)

Second Case: Boys at B, Girls at A
Number of ways for boys to seat on bench B = 4!
Number of ways for girls to seat on bench A = 4!
Second Case = (4!)(4!)

Total number of ways = First Case + Second Case

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