In how many ways can 3 boys and 3 girls be seated on 8 chairs so that no two girls are together?

Let us take opposite of the constraint.
2 girls sitting together: :
1 case is GGBGBBBB.
Total number of ways=3!*5!*5 with just shifting the rightmost girl.
Then the 2 leftmost girls can shift one position , and using the above reasoning, the total number of ways = 3!*5!*4 and so on till the rightmost girl has 1 position.
So total number of ways = 3!*5!(5+4+3+2+1)=120*90=10800
Similarly another case is:
GBGGBBBB.
Using the above reasoning, the total number of cases is: 3!*5!*(15) =10800
Let us take 3 girls sitting together
GGGBBBBB
There are 3! *5! Ways. The 3 leftmost girls can shift 6 positions. So there are a total of 3!*5!*6=4320 ways
So total is 2*10800 + 4320=25920
The total number of possibilities = 8! Ways =40,320
Answer is 40320-25920=480
Hence C.