In how many ways can one select eight integers $a_1,a_2, ... ,a_8$, not necesarily distinct, such that $1 \le  a_1 \le ... \le a_8 \le 8$?
1. We need to find the number of ways to select eight integers \(a_1, a_2, \ldots, a_8\) such that \(1 \le a_1 \le a_2 \le \cdots \le a_8 \le 8\). This is a classic combinatorial problem that can be solved using the "stars and bars" theorem.

2. The "stars and bars" theorem states that the number of ways to distribute \(n\) indistinguishable objects into \(k\) distinguishable bins is given by \(\binom{n+k-1}{k-1}\).

3. In this problem, we can think of the integers \(a_1, a_2, \ldots, a_8\) as 8 indistinguishable stars that need to be placed into 8 distinguishable bins (the integers from 1 to 8). The condition \(1 \le a_1 \le a_2 \le \cdots \le a_8 \le 8\) ensures that the stars are placed in non-decreasing order.

4. To apply the "stars and bars" theorem, we need to find the number of ways to place 8 stars into 8 bins. This is equivalent to finding the number of non-negative integer solutions to the equation:
   \[
   x_1 + x_2 + \cdots + x_8 = 8
   \]
   where \(x_i\) represents the number of stars in the \(i\)-th bin.

5. According to the "stars and bars" theorem, the number of solutions is given by:
   \[
   \binom{8+8-1}{8-1} = \binom{15}{7}
   \]

6. Therefore, the number of ways to select eight integers \(a_1, a_2, \ldots, a_8\) such that \(1 \le a_1 \le a_2 \le \cdots \le a_8 \le 8\) is \(\boxed{\binom{15}{7}}\).