3 girls and 3 boys are being arranged in a line such that girls and boys alternate. What is the number of arrangements possible

So the way I am trying to solve this is as follows:

• Allow the girls to be arranged first : $3!$
• There are 4 spaces that are created for the three boys : $4P3$ (since the order is important $4C3 \times 3!$, which is $4P3$)

Therefore total possibilities $=3! \times 4P3 = 144$

Now the answer to the problem is $72$, which is half of my answer. What am I double counting?

• What is "a line"? If you reverse the line, is that different or not? – GEdgar May 18 '18 at 12:45

If you use your method, than these lines are also accepted too:

$$-G-G-G-\Rightarrow BG-GBGB;BGBG-GB$$

For each of the two lines above, there are $3!$ ways to rearrange boys and $3!$ ways to rearrange girls, so you have include $2\times 3!\times 3!=72$ incorrect cases ("$-$" in the line above stands for "empty").

Girls and boys are counted seperately.

For each case (girls first and boys first are two distinct cases), the number of ways to rearrange girls is $3!$, the number of ways to rearrange boys is also $3!$.

So the answer should be $3!\times\ 3!\times 2=72$.

• I see. So I dont have 4 spaces available, I technically only have 3 spaces available, but two separate ways of arranging that 3 spaces – Dave T May 18 '18 at 12:50
• I think you mean "two ways to choose three spaces". By the way, welcome to MSE, you might want to check this tour page for a quick overview of how the site works. – user061703 May 18 '18 at 12:54

An option:

1) B-G-B-G-B-G,

i.e start the chain with a boy.

$3! ×3!$, i.e $3!$ ways to arrange the boys times 3! ways to arrange the girls in the above chain.

2) G-B-G-B-G-B, i.e. start the chain with a girl.

Similar argument as in 1): $3!×3!$ ways.

3) Altogether: $2(3!×3!)$ ways.