In how many ways can the $6$ letters in the list $G,H,I,J,K,L$ be rearranged so that $G$ is the third letter in the list and $H$ is not next to $G$?

Question

The $6$ letters in the list $G,H,I,J,K,L$ are to be rearranged so that $G$ is the third letter in the list and $H$ is not next to $G$. How much such arrangements are possible?

I'm guessing the answer is $72=3 \cdot (4 \cdot 3 \cdot 2 \cdot 1)$. Is this right?

Answer 1

You are correct. We can confirm your answer by using complementary counting.

Since the position of G is fixed, the remaining five letters can be placed in the remaining five positions in $5!$ ways. From these, we subtract those arrangements in which H is next to G. Since G is in the third position, there are two ways to place H next to G. Once H has been placed, the remaining letters can be placed in $4!$ ways. Hence, there are $2 \cdot 4!$ prohibited arrangements. Thus, the number of arrangements in which G is in the third position and H is not next to G is $5! - 2 \cdot 4! = 120 - 48 = 72$, as you found.

Answer 2

You are correct! G is fixed so we don't have to worry about that. Then there are 3 places for H to go, for each of those places there are 4! ways to place the remaining letters so there are 3*(4!) = 72 rearrangements.