Question

Five boys and-five girls form a line with the boys and girls alternating. Find the number of ways of making the line. In how many different ways could they form a circle such that the boys and girls alternate?

Answer 1

5 boys can be arranged in a line in 5! ways. Now here the boys and girls are alternating. Let the boys be
$B_1\times B_2\times B_3\times B_4\times B_5$
We are left with 5 places marked by cross where five girls can be arranged in 5!ways. Thus by fundamental theorem it will-be 5! $\times$ 5! = $(5!{)}^2$
Now the position (1) could be
$\times B_1\times B_2\times B_3\times B_4\times B_5$
We are left with 5 places marked by cross where five girls can-be arranged in 5! ways. Thus by fundamental theorem it will be 5! $\times$ 5! = $(5!{)}^2$ Hence the total number of ways will be $(5!{)}^2+(5!{)}^2=2(5!{)}^2$
In other words we can say if we start the line with the boy in the 1st position as in (1) there are $(5!{)}^2$ ways& if we start with girl in 1st position as in (2) then also there are $(5!{)}^2$ ways. Thus the total number of ways will be. Thus the total number of ways will be
$(5!{)}^2+(5!{)}^2=2(5!{)}^2$
Had the condition been that no to girls are together instead of boys and girls alternating then we would have followed gap method as in Q. 18 (a), R 328-332
After fixing the boys.
$\times B_1\times B_2\times B_3\times B_4\times B_5\times$
We have six places marked by cross where .5 girls can .be arranged in, ${}^6{P}_5=\frac{6!}{1!}=6!$ ways and .hence the answer in this case would be 5! .6!. In this case two boys can sit together because you have to ignore one cross say between $B_3$ and $B_4$ and these two boys will be together.