There are 4 boys and 4 girls. They want to sitting in a row. How many ways to sit, if given conditions there is no 2 boys or 2 girls sitting next to each other?
To answer this question, I use complement rule.
Number ways to sit if no condition: $8!$.
Now I confused to determine number ways to sit if 2 boys and 2 girls sitting next to each other. Is it right we have to arrange
(B1,B2)(B3,B4)(G1,G2)(G3,G4)?
Number of ways to arrange B1,B2,B3,B4 and G1,G2,G3,G4 is $4!\times 4!$
Number of ways to arrange (B1,B2)(B3,B4)(G1,G2)(G3,G4) is $4!$
So we have number ways to sit if 2 boys and 2 girls sitting next to each other is $(4!)^3$.
So, the answer is $8!-(4!)^3$. Is it right answer?
Best Answer
If no pair of boys or girls sit next to each other, then they must alternate.
Pick which gender sits in the leftmost seat $(2)$. Then choose the order of the four boys $(4!)$ and the order of the four girls $(4!)$.
So $2 \cdot 4!^2 = 1152$ ways.