How many ways are there to permute the 8 letters A, B, C, D, E, F, G, and H so that A is not at the beginning and H is not at the end?

There are 8 letters.

We want an arrangement which does not begin with A and does not end with H.

First, let us determine the number of possible choices for the first and last letters.

First letter.

The number of choices available for the first letter is 7. The reason is that, out of the 8 letters only letter A cannot occupy this position. Thus, we got 7 choices for this position.

Last letter

The number of choices available for the last letter is 6. The reason is that for this position, we exclude the letter H and the letter occupying the first position. Thus the number of choices is $8-2=6$

The number of permutations for the first and last letters is $7\times 6=42$ permutations.

Between the first and the last letters

Excluding the letters in the first and last positions, we remain with $8-2=6$ letters.

Now,

In the second position, we can choose from any of the 6 letters thus, there are 6 ways to choose the six letters to occupy the second position

For the third position, there are $8-3=5$ letters to choose from to occupy this position thus there are 5 ways to choose the 5 letters to occupy the third position.

For the fourth position, there are $8-4=4$ letters to choose from to occupy this position thus there are 4 ways to choose the 4 letters to occupy the fourth position.

For the fifth position, there are $8-5=3$ letters to choose from to occupy this position thus there are 3 ways to choose the 3 letters to occupy the fifth position.

For the sixth position, there are $8-6=2$ letters to choose from to occupy this position thus there are 2 ways to choose the 2 letters to occupy the sixth position.

For the seventh position, there is $8-7=1$ letter remaining to occupy this position thus there is only 1 way for the remaining letter to occupy this position.

Now, the number of permutations for the second to the seventh letter (between the first and the last letter) is $6\times5\times4\times3\times2\times1=720$ permutations.

Therefore the total number of permutations is $720\times 42=30240$ permutations.