In how many ways can 3 boys and 3 girls be seated on 8 chairs so that no two girls are together?	Choose the correct answer.
A) 5760
B) 14400
C) 480
D) 56
E) 40320
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= 
<gadget id="calculator">120*90</gadget>
<output>10_800</output>
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= 
<gadget id="calculator">  2*10800 + 4320</gadget>
<output>25_920</output>
25920   The total number of possibilities= 8! Ways =40,320   Answer is 40320-25920=480   Hence 480.
<result>C</result>
In how many ways can 3 boys and 3 girls be seated on 8 chairs so that no two girls are together?