7. Between 2020 and 2400, there are ( ) integers whose four digits are strictly increasing and all different (for example, 2357 is one such integer).
(A) 9
(B) 10
(C) 15
(D) 20
(E) 28
7. C.

Since the digits are strictly increasing, the second digit cannot be 1 or 2.

Also, the four-digit number cannot be greater than 2400, so the second digit cannot be 4. Therefore, the second digit can only be 3. Hence, the last two digits can only be chosen from the set $\{4,5,6,7,8,9\}$. Thus, there are $\mathrm{C}_{6}^{2}=\frac{6 \times 5}{1 \times 2}=15$ in total.