18) The sum of all two-digit numbers with distinct digits is:
(A) 3840
(B) 3960,
(C) 4140,
(D) 4260,
(E) 4410.
18. The answer is $(\mathbf{E})$.

The sum of all two-digit numbers is

$10+11+\cdots+98+99=(1+2+\cdots+98+99)-(1+2+\cdots+8+9)=\frac{99 \cdot 100}{2}-45=45 \cdot 109$.

From this sum, I need to subtract the sum of the numbers formed by two identical digits:

$$
11+22+\cdots+88+99=11(1+2+\cdots+8+9)=11 \cdot 45 .
$$

The result is therefore $45 \cdot 109-45 \cdot 11=45 \cdot 98=4410$.