5. In decimal, if a positive integer with at least two digits has all its digits, except the leftmost one, smaller than the digit to their left, then it is called a descending positive integer. The total number of such descending positive integers is ( ).
(A) 1001
(B) 1010
(C) 1011
(D) 1013
5.(D).

Let the leftmost digit be $a$, then the digits $a-1$, $a-2, \cdots, 2,1,0$ that are smaller than $a$ can each appear at most once to the right of $a$. Therefore, the number of decreasing positive integers starting with $a$ is equal to the number of non-empty subsets of $\{a-1, a-2, \cdots, 2,1,0\}$, which is $2^{a}-1$. Since $a=1,2,3, \cdots, 9$, the total number of decreasing positive integers is
$$
\sum_{a=1}^{9}\left(2^{a}-1\right)=\left(\sum_{a=1}^{9} 2^{a}\right)-9=2^{10}-2-9=1013 \text { (numbers). }
$$