Six friends go to see a comedy at the local movie theater and sit in a particular row together. If Grady and Howard are among those 6 friends but can never sit next to each other, how many 6-person seating arrangements are possible? Pick one
A) 360	B) 420	C) 480	D) 600	E) 720
Ways to arrange 6 people in a row = 6! = 720   From this we want to subtract all the arrangements where Grady and Howard are next to each other. If we consider Grady and Howard one unit (GH), then the number of arrangements of the GH unit and the 4 other friends is 5! And the number of arrangements of G and H within the GH unit are 2 (GH or HG). So the total number of arrangements of the 6 friends with Grady and Howard together are 5!*2   Number of permutations where Grady and Howard are not seated next to each other = 6! - 5!*2 = 720-120*2= 
<gadget id="calculator">720-120*2</gadget>
<output>480</output>
480   Answer: 480
<result>C</result>
Six friends go to see a comedy at the local movie theater and sit in a particular row together. If Grady and Howard are among those 6 friends but can never sit next to each other, how many 6-person seating arrangements are possible?