0 votes 0 votes The number of ways in which 4 boys and 3 girls can be seated in a row such that girls and boys are alternate is :: Answer given by ME is $4!*3! = 144$, but i think $\binom 5 3*4!*3!$ Should be correct. thor asked Jan 7, 2017 thor 601 views answer comment Share Follow See 1 comment See all 1 1 comment reply Uzumaki Naruto commented Jan 7, 2017 reply Follow Share For better clarity, instead of fixing boys(as it says they are alternate), fix the positions of girls as _G_G_G_ which is 3! * 4! 1 votes 1 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes First Boys are arranged in 4! ways now we have 3 positions for girls So, 3C3 B_B_B_B Now girls can be swap their seats therefore 3C3* 3! so total , 3!*4! utk0203 answered Jan 7, 2017 selected Jan 7, 2017 by thor utk0203 comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments Devwritt commented Jan 7, 2017 reply Follow Share @thor __B1__B2__B3__B4__ five positions for Girls. Ths is wrong approach because if all 3 girls seats in first 3 position , then arrangement will be like: G1 B1 G2 B2 G3 B3 B4 Last two boys violates the condition . So answer is : 3! * 4! = 144 1 votes 1 votes thor commented Jan 8, 2017 reply Follow Share thanx!! 0 votes 0 votes thor commented Jan 8, 2017 reply Follow Share So, can i say The number of ways in which n boys and m girls can be seated in a row such that girls and boys are alternate is $0$ if | n -m | > 1 ? 0 votes 0 votes Please log in or register to add a comment.