Given: There are 5 boys and 3 girls that can be seated in a row such that no two girls are together. First let's calculate number of ways to arrange boys in the row. ∵ There are 5 boys so they will occupy 5 places in 5P5 = 5! ways = 120 ways We can arrange the 3 girls in 6 places shown by cross mark : X B X B X B X B X B X Number of ways in which 3 girls can seat in 6 places denoted by cross mark =6P3 As we know that, nPr=
n!
(n−r)!
⇒6P3=120 ∴ The required number of ways =120×120=14400