How many 5 digit positive integers are there such that each of its digits, except for the last one, is greater than or equal to the next digit?
1. We need to find the number of 5-digit positive integers such that each of its digits, except for the last one, is greater than or equal to the next digit. Let the 5-digit number be represented as \(abcde\), where \(a, b, c, d, e\) are its digits.

2. The condition given is that each digit, except for the last one, is greater than or equal to the next digit. This can be written as:
   \[
   a \geq b \geq c \geq d \geq e
   \]
   where \(a, b, c, d, e\) are digits from 0 to 9.

3. To simplify the problem, we can use a combinatorial approach. We need to count the number of non-increasing sequences of length 5 where each digit is between 0 and 9.

4. We can transform the problem into a problem of distributing indistinguishable objects (stars) into distinguishable bins (digits). This is equivalent to finding the number of solutions to the equation:
   \[
   x_1 + x_2 + x_3 + x_4 + x_5 = 9
   \]
   where \(x_i\) represents the number of times digit \(i\) appears in the sequence, and \(x_i \geq 0\).

5. The number of non-negative integer solutions to this equation is given by the stars and bars theorem, which states that the number of solutions to the equation \(x_1 + x_2 + \cdots + x_k = n\) is:
   \[
   \binom{n+k-1}{k-1}
   \]
   In our case, \(n = 9\) and \(k = 5\), so we need to find:
   \[
   \binom{9+5-1}{5-1} = \binom{13}{4}
   \]

6. Calculating the binomial coefficient:
   \[
   \binom{13}{4} = \frac{13!}{4!(13-4)!} = \frac{13!}{4! \cdot 9!} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 715
   \]

Therefore, the number of 5-digit positive integers such that each of its digits, except for the last one, is greater than or equal to the next digit is \( \boxed{715} \).