Problem: How many even integers between 4000 and 7000 have four different digits? Solution: We do casework by counting the numbers starting with $4,5,6.$ In the case of $4$ and $6$ being the starting digits, we have $4$ choices of even digits for the unit digit. From here, we have $8$ choices for the second digit and $7$ choices for the third digit, so these cases contribute $2 \cdot (4 \cdot 7\cdot 8)$ numbers. When the number starts with $5$, there are now $5$ choices of even digits for the unit digit, followed by $8$ choices for the second digit and $7$ choices for the third digit, contributing $5 \cdot 7 \cdot 8$. Summing, we get $13 \cdot 7 \cdot 8 = 728$.