# If 4 boys and 4 girls line up completely at random, what is the probability that the 4 boys are together and the 4 girls are together?

The question requires the lineup to be either BBBBGGGG or GGGGBBBB. I'm not sure if my thinking is correct, but this is my work so far:

Boys (notated as $$B_1$$ to $$B_4$$):

The first boy has 8 spots to stand in, therefore $$P(B_1)$$= $$\frac 18$$.

The second boy has 3 spots to stand in, therefore $$P(B_2)$$= $$\frac 13$$.

The third boy has 2 spots to stand in, therefore $$P(B_3)$$= $$\frac 12$$.

The fourth boy has only 1 spot to stand in, therefore $$P(B_4)$$= 1.

Girls (notated as $$G_1$$ to $$G_4$$):

The first girl only has 4 spots to stand in since the 4 boys occupied 4 spots already, therefore $$P(G_1)$$= $$\frac 14$$.

The second girl then has 3 spots to stand in, therefore $$P(G_2)$$= $$\frac 13$$.

The third girl has 2 spots to stand in, therefore $$P(G_3)$$= $$\frac 12$$.

The fourth girl has only 1 spot to stand in, therefore $$P(G_4)$$= 1.

Then $$P(BBBBGGGG)$$= ($$\frac 18$$ $$\cdot$$ $$\frac 13$$ $$\cdot\frac 12$$ $$\cdot$$ $$1$$ $$\cdot$$ $$\frac 14$$ $$\cdot$$ $$\frac 13$$ $$\cdot$$ $$\frac 12$$ $$\cdot$$ $$1$$) = $$.000868$$, which is the same for $$P(GGGGBBBB)$$, therefore the final answer is $$0.001736$$.

Is this correct or did I misunderstand anything?

• Not sure of your notation. What is $P(B_1)$? In any case, any arrangement of the kids is determined by the locations of the girls...tho place the girls you need to pick four seats out of eight, so there are $\binom 84=70$ arrangements. Only two work for you, so the answer is $\frac 2{70}\approx 0.028571429$. – lulu Sep 29 '18 at 18:52
• @lulu By $\frac{2}{70}$, do you mean that only 2 arrangements will give all 4 girls together? But how about the boys? – peco Sep 29 '18 at 19:01
• No...I meant what you said. There are exactly two arrangements in which all the boys AND all the girls are together. – lulu Sep 29 '18 at 19:04
• I think I understand now, thanks! – peco Sep 29 '18 at 19:05

There are $$2$$ cases: $$B_1B_2B_3B_4G_1G_2G_3G_4$$ or $$G_1G_2G_3G_4B_1B_2B_3B_4$$,and the probability for each is $$\dfrac{4!\times 4!}{8!}$$. Thus the probability boys together and girls together is: $$\dfrac{2\times 4!\times4!}{8!}$$
• Just to confirm, having $\frac{4!}{8!}$ will be the probability of all 4 boys, for example, together? – peco Sep 29 '18 at 19:06
• @peco: no you have to account for all positions. You also have to account for the girls' order, even when do not stand together. So it is actually $5 \times \frac{4!4!}{8!}$ – Alex Sep 29 '18 at 20:00