counting question

To solve this problem, we can use the principle of inclusion and exclusion.

Let A be the set of all ways to distribute 8 indistinguishable balls among 15 distinguishable boxes.

Let B_i be the set of all ways to distribute the balls such that box i is empty.

We want to count the number of ways to distribute the balls such that at least 3 boxes are empty. This is the same as counting the number of ways to distribute the balls such that 0, 1, or 2 boxes are empty.

We can use the following formula:

|A - B_1 - B_2 - ... - B_n| = |A| - |B_1| - |B_2| - ... - |B_n| + |B_1 \cap B_2| + |B_1 \cap B_3| + ... + |B_{n-1} \cap B_n| - |B_1 \cap B_2 \cap B_3| - ... - |B_{n-2} \cap B_{n-1} \cap B_n| + ... + |B_1 \cap B_2 \cap B_3 \cap ... \cap B_n|

where |A| is the cardinality of set A, and |B_i| is the cardinality of set B_i.

In our case, we have:

|A| = \binom{15}{8} = 6435

|B_i| = \binom{14}{7} = 3432

|B_1 \cap B_2| = \binom{13}{6} = 1771

|B_1 \cap B_3| = |B_2 \cap B_3| = \binom{12}{5} = 792

|B_1 \cap B_2 \cap B_3| = \binom{11}{4} = 330

Therefore, the number of ways to distribute the balls such that at least 3 boxes are empty is:

6435 - 3 * 3432 + 3 * 1771 - 3 * 792 + 1 * 330 = \boxed{1818}