Q) No of ways in which $6$ boys and $6$ girls can be arranged so that no $2$ boys or girls are together.
My Approach
B_B_B_B_B_B_ The boys can arrange themselves in $6!$ ways same for the girls. Hence there are $6!*6!$ ways. But this is not the answer. What did I do wrong?
On
I have the following idea: Let us arrange the students in this way: $$CDC_1DC_2DC_3DC_4DC_5DC$$ where $C$ represents the places that can be taken by students of one type(boys$\mid$girls) and $D$ by te students of brother type(girls$\mid$boys), i.e., if $C$ is first occupied by the girls and $D$ by the boys, the next arrangement should be vice versa.
We have $7 C$'s but there are only six students to take their places. But we cannot arrange them anywhere because then that will result in $2 D$'s coming together. Hence, they should definitely occupy all places $C_i \in [1,5]$. Hence, the sixth person can occupy one of the two unmarked $C$'s. So the total number of ways are $2$ ways.
We can arrange the six students in $\binom{6}{5}\times 5! =720$ ways.
Also, the $6 D$'s can be arranged in $6! = 720$ ways.
The total ways for the seating is: $2\times 720\times 720$ ways.
The answer is thus: $$2\times (6!)^2$$ Hope it helps.
You can start with a girl instead of a boy, so there are twice as many patterns as you compute.