Problem Solving

OA coming after some people answer:

5 girls and 3 boys are arranged randomly in a row. Find the probability that:

A) there is one boy on each end.

B) There is one girl on each end.

Total no: of ways to arrange them = 8!

No: of ways in which boys are the end

we can select 2 boys from 3 in 3 ways ...

and arrange them in 2 ways

we can arrange the remaining 6 in 6! ways


So the total no of ways = 2*3*6!


the probability = 6*6!/8!= 6/7*8=3/28


Similarly no of ways in which gals are in the end = 5c2*2*6!
=20*6!
The probability= 20*6!/8!=20/56=5/14

There can be mistakes ...

I get the same answer.

1. 3P2 x 6!/8! = 3/28

2. 5P2 x 6!/8! = 5/14

OA:

For the first scenario:
A) there is one boy on each end.

The first seat can be filled in 3C1 (3 boys 1 seat) ways = 3
the last seat can be filled in 2C1 (2 boys 1 seat) ways = 2
the six seats in the middle can be filled in 6! (1 boy and 5 girls) ways
Total possible outcome = 8!
Probability= (3C1 * 2C1 * 6!)/ 8! = 3/28

For the second scenario:
A) there is one girl on each end.

The first seat can be filled in 5C1 (5 girls 1 seat) ways = 5
the last seat can be filled in 4C1 (2 girls 1 seat) ways = 4
the six seats in the middle can be filled in 6! (3 boys and 3 girls) ways
Total possible outcome = 8!
Probability= (5C1 * 4C1 * 6!)/ 8! = 5/14