Question 364413
Let's assume the 4 girls are G1, G2, G3, G4 and the 4 boys B1, B2, B3, B4. 
Let G1,B1 be the pair that needs to sit always together

As per problem, the boys and girls should sit alternatively.

The possibile scenarios are:

G1B1  G2B2G3B3G4B4  (girl, boy, girl, boy....)

B1G1  B2G2B3G2B4G4  (boy, girl, boy, girl....)

In the 1st arrangement, G2,G3 and G4 can be arranged in 3! ways and B2,B3 and B4 can be arranged in 3! ways

Hence, total no of ways the 3 girls and 3 boys can be arranged is 3! * 3! = 36 ways

The G1B1 doublet can be placed in any of the 7 positions (4 positions as G1B1 and 3 positions as B1G1 satisfying the condition that boys and girls should sit alternatively)  

So, the number of arrangements = 7*36 = 252

We have another set of 252 arrangements for the 2nd scenario 

B1G1  B2G2B3G2B4G4  (boy, girl, boy, girl....)

 So, total no of ways = 252 + 252 = 504 ways