How many possible arrangements of the word "HAIRCUT" such that one letter is in the right spot?

Question

My thought process:

I know that to have no letters in their correct spots, I need to use the derangement formula, leaving me with:

$$8!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}+\frac{1}{8!}\right)$$

If that is the total amount, I know I would need to subtract the possible scenarios of one letter being in the right spot, but I cannot figure out what to do.

My question is: What am I supposed to subtract by to give me the possible arrangements with only one letter in the correct spot? Thank you.

Answer 1

Hint: Choose which letter is in the right spot. You then have a derangement on the remaining six letters.

Answer 2

Choose 1 letter to be in the right spot, there are 7 choices of letter H, A, I, R, C, U, and T. Then dearrange the rest of the 6 words, without changing the condition of the letter we choose, there are $!6=265$ possible ways. The total number of ways would be$$7\cdot !6=1855$$