Probability of boys ahead girls

Question

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..

Answer 1

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 $1-\left(\frac25 +\frac1{10}\right) = \frac12$

Answer 2

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 $60/120=1/2$.