Thanks Jasonkiln,

You are right of course, I cannot just divide by 2 at the end

.I can tie my two Cs together though. becasue any answer with 2 Cs tied together will not be included and I will have to subtract.

So I have

aauryC and I will add the other c later

6!/2! = 360

Now where can the last c go

|a|a|u|r|y|C

It can go where any of those bars are. It can't go after the C becasue that would be the same as before the C Cc = cC

There are 6 bars.

360*6=2160

There are 2160 permutations where 2 or more Cs are together

3360-2160 = 1200

There are 1200 permutations where no 2 Cs are together.

This is now the same as Ginger's answer, Builderboi agrees too. I think his method is similar to mine.

Ginger's method is a little more simple.