Combinatorics - How many numbers with all different digits are between $1000$ and $7856$?

Question

To answer this question I was trying to solve the problem by parts. So, first I counted from 1000 to 7000: $6*9*8*7$ that means there are $6$ possibilities on position one, $9$ to position 2, $8$ to position 3 and $7$ for the last one. Then, I counted from 7000 to 7800 and I got this: $7*8*7$. In tens, I counted from 7800 to 7850 and I think this part is wrong but I got this: $4*7$, when there are 4 possibilities on tens and 7 possibilities on units because I need to put away 3 digits above. Finally, on units I only got 5, so the all sum is: $6*9*8*7+7*8*7+4*7+5=3449$. I want to know if I solve this problem well. I'm learning English, so I'm sorry if you can't understand me.