in how many ways can 9 men stand in a row if A and B can not stand side by side, and Cmust stand at the end.?

Question

I have come up with :

ADBEFGHIC = 6 different ways
ADEBFGHIC =5 different ways
ADEFBGHIC = 4 different ways
the pattern led me to 6! then the same would be if I switched A and B's position...

I came up with the answer 6!*6!

it feels like I'm missing something. Could you tell me if there is more work to the problem?

Thank you

Answer 1

C is at the end, so its just the other 8 positions to play with.

Count how many permutations of those 8 there are.

Then count how many there are where A and B stand side by side.

Then subtract those out.

Edit: Poster below me is way off. without any restrictions, 9 men could be arranged in over 360000 ways. Not 81.

Basically, you can take 'C' out of it. There are 8 spots left, and 8 men. Spot 1 can hold any of the 8. spot 2 any of the remaining 7 and so on, giving 8X7X6X5X4X3X2 which is 8! possibilities.

That's 40320.

Of those 40320, some involve A and B being side by side. How many? Well, they could be in slot 1&2, 2& 3 .... up to 7&8. And they could be A-B or B-A. That's 14 possibles. For each of those 14, the other 6 spots could be filled 6! ways.

So there are 14 X 6! = 14 X 720 = 10800 combos with C in the 9 spot and A & B side by side.

Therefor, of the 40320 with C in the 9 spot, 40320-10800 = 30240 possible arrangements have C at the end, but A and B NOT side by side.

Answer 2

You are way off. Hint: 9 men could stand in a row 81 different ways without your restrictions added. Work from that.