Probability of boys ahead girls There are three boys and two girls in a Queue. What is the Probability that number of boys ahead of every girl is at least one more than number of girls ahead of her.
My Try: Let $B1$,$B2$,$B3$ are Boys and $G1$,$G2$ are Girls
The Possible Outcomes are:
$1.$ $G1$ $G2$ $B1$ $B2$ $B3$ 
$2.$ $G1$ $B1$ $G2$ $B2$ $B3$
$3.$ $G1$ $B1$ $B2$ $G3$ $B3$
$4.$ $B1$ $G1$ $G2$ $B2$ $B3$
$5.$ $B1$ $G1$ $B2$ $G2$ $B3$
So The Required Probability is $$\frac{5\times2!\times3!}{5!}=\frac{1}{2}$$ Please let me know if there is better way to do this..
 A: Not much quicker, but you could look at patterns that fail
. . . . G with prob 2/5
B B G G B with prob 1/10

and $1-\left(\frac25 +\frac1{10}\right) = \frac12$
A: Consider the places 54321, the queue's direction is from right to left. There are $5!=120$ total arrangements. To satisfy the given condition for both the girls simultaneously, first position should be occupied by a boy only. To find the no. of arrangements where a boy has first position we should find the no. of arrangements where a girl has first position..i.e $2(4P_1\times3P_3) 
                                                                    =2(4x6)   
                                                                       =48$
Therefore, no. of arrangements where boys have first position is $120-48 = 72$. In these 72 arrangements, also girls cant have 2nd and 3rd positions simultaneously. Hence, the number of remaining arrangements is $72-2 \times 3P_3
             =72-12
             =60$. Therefore the probability is $60/120=1/2$.
