How many different arrangements are possible of the letters of the word CARLGAUSS if no S can be in the middle (5th) position?

Question

What I want to do is to first calculate all the possible permutations of the letters of the given word. Once I do that, I plan to keep an S in the 5th position and calculate possible permutations. But the question is do I have to multiply it by 2 and THEN deduct it from the total number of perms? Or will I get the correct answer if I directly deduct it? Thank you.

Answer 1

Total number of permutations = (number of letters)!/(repeated letters)! = (9!)/(2!2!) = 362880/4 = 90720

the repeating letters are A(2 times) and S(2 times).

Number of permutations with S in the 5th position = (number of letters - 1)/(repeated letters)! = 8!/2! = 20160

the repeating letter now is only A(2 times), one S is reserved in the 5th position so there will be no repetitions of S.

Number of permutations without S in the 5th possition = Total number of permutations - Number of permutations with S in the 5th position = 90720 - 20160 = 70560

Answer 2

Assign each letter a number $1-9$ (ignore duplicates for the moment).

Then there are $9!=362,880$ permutations, of which $8!=40320$ have an $8$ in the middle, and $8!$ have a $9$ in the middle.

So we are left with $9!-8!-8!=282,240$ cases. These are duplicated with A and S, so we divide by $4$ to give $70,560$.