How many $6$-digit positive integers have their digits in nondecreasing order from left to right? Note that $0$ cannot be a leading digit.
To determine how many $6$-digit positive integers have their digits in nondecreasing order from left to right, we can use the combinatorial method known as "stars and bars."

1. **Define the problem in terms of combinatorics:**
   We need to find the number of ways to distribute $6$ digits (from $1$ to $9$) such that the digits are in nondecreasing order. This is equivalent to finding the number of nonnegative integer solutions to the equation:
   \[
   x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 = 6
   \]
   where $x_i$ represents the number of times digit $i$ appears in the $6$-digit number.

2. **Apply the stars and bars theorem:**
   The stars and bars theorem states that the number of nonnegative integer solutions to the equation:
   \[
   x_1 + x_2 + \cdots + x_k = n
   \]
   is given by:
   \[
   \binom{n+k-1}{k-1}
   \]
   In our case, $n = 6$ (the total number of digits) and $k = 9$ (the digits from $1$ to $9$).

3. **Calculate the binomial coefficient:**
   Using the stars and bars formula, we get:
   \[
   \binom{6+9-1}{9-1} = \binom{14}{8}
   \]
   We can simplify this using the symmetry property of binomial coefficients, $\binom{n}{k} = \binom{n}{n-k}$:
   \[
   \binom{14}{8} = \binom{14}{6}
   \]

4. **Compute the binomial coefficient:**
   \[
   \binom{14}{6} = \frac{14!}{6!(14-6)!} = \frac{14!}{6! \cdot 8!}
   \]
   Calculating the factorials, we get:
   \[
   14! = 87,178,291,200
   \]
   \[
   6! = 720
   \]
   \[
   8! = 40,320
   \]
   Therefore:
   \[
   \binom{14}{6} = \frac{87,178,291,200}{720 \cdot 40,320} = \frac{87,178,291,200}{29,030,400} = 3003
   \]

Thus, the number of $6$-digit positive integers with digits in nondecreasing order is $\boxed{3003}$.