5 boys and 5 girls are sitting in a row randomly. The probability that boys and girls sit alternately is...... ?

There are total 10 seats. _ _ _ _ _ _ _ _ _ _. Case 1 1st seat is occupied by a boy. For this, the number of choices are 5. 2nd seat is occupied by a girl. For this number of choices again are 5 3rd seat is occupied by a boy. For this, the number of choices are 4. 4th seat is occupied by a girl. For this number of choices again are 4 5th seat is occupied by a boy. For this, the number of choices are 3. 6th seat is occupied by a girl. For this number of choices again are 3 7th seat is occupied by a boy. For this, the number of choices are 2. 8th seat is occupied by a girl. For this number of choices again are 2 9th seat is occupied by a boy. For this, there is 1 choice only 10th seat is occupied by a girl. For this, there is 1 choice only. So total number of choices for Case 1 = 5*5*4*4*3*3*2*2*1*1 we can rearrange it and write this as = 5 *4*3*2*1*5*4*3*2*1 = 5!*5!. Case 2 Here we start with girl sitting in the first seat. Similar to the first case here also total ways of arranging the students alternately would be = 5!*5! Therefore total no. of ways = 2* 5!*5! Sample space [all arrangements] is = 10! as there are 10 seats. So, the probability of arranging them in alternate positions = 2* 5!*5! /10! by cancelling one 5! from numerator and denominator we get, = 2 * 5!/ (10*9*8*7*6) = 2 * 5*4*3*2*1/ (10*9*8*7*6) = 1/(9*2*7) = 1/126 read less

After placing all the 5 boys, the girls can be placed in between them, and there are 6 places, as indicated by * as given below: *b*b*b*b*b* So, 5 boys can be arranged among themselves in 5! ways, and the girls can be arranged in possible 6 places in 6p5 ways=6!. So the total number of required arrangements is 5!6!. Total number of possible arrangements (without any restriction) is 10!. So the required probability is 5!6!/10! = 1/42 read less