In how many ways can you distribute $8$ indistinguishable balls among $6$ distinguishable boxes, if at least three of the boxes must be empty?

Answer 1

To solve the problem of distributing \(8\) indistinguishable balls among \(6\) distinguishable boxes with the condition that at least three of the boxes must be empty, we can start by calculating the total possible distributions of the balls and then account for the condition on the boxes.

1. Determine the number of non-empty boxes: Since at least 3 boxes must be empty, it follows that at most \(6 - 3 = 3\) boxes can be non-empty. Let \(k\) be the number of non-empty boxes. Thus, \(k\) can take the values \(1, 2, 3\).

2. Calculate distributions for each case based on \(k\).

   Case 1: \(k = 1\) (1 non-empty box):

   • We need to put all \(8\) balls into \(1\) box. There is only \(1\) way to do this.
   • Thus, the number of distributions for \(k = 1\) is \( \binom{6}{1} \cdot 1 = 6\).

   Case 2: \(k = 2\) (2 non-empty boxes):

   • We need to distribute \(8\) balls into \(2\) boxes. The formula for distributing \(n\) indistinguishable balls into \(k\) distinguishable boxes is given by the stars and bars theorem as \( \binom{n+k-1}{k-1} \).
   • Here, \(n = 8\) and \(k = 2\): \[ \text{Number of ways} = \binom{8 + 2 - 1}{2 - 1} = \binom{9}{1} = 9 \]
   • The number of ways to choose which \(2\) boxes will be non-empty is \(\binom{6}{2}\).
   • Thus, the total for \(k = 2\) is: \[ \binom{6}{2} \cdot 9 = 15 \cdot 9 = 135. \]

   Case 3: \(k = 3\) (3 non-empty boxes):

   • We distribute \(8\) indistinguishable balls into \(3\) boxes. Using the same stars and bars theorem: \[ \text{Number of ways} = \binom{8 + 3 - 1}{3 - 1} = \binom{10}{2} = 45. \]
   • The number of ways to choose which \(3\) boxes will be non-empty is \(\binom{6}{3}\).
   • Thus, the total for \(k = 3\) is: \[ \binom{6}{3} \cdot 45 = 20 \cdot 45 = 900. \]

3. Combine all cases: The total number of ways to distribute \(8\) indistinguishable balls into \(6\) distinguishable boxes, ensuring at least \(3\) boxes are empty, is calculated by adding the results from all cases: \[ 6 \text{ (for } k=1\text{)} + 135 \text{ (for } k=2\text{)} + 900 \text{ (for } k=3\text{)} = 1041. \]

Thus, the total number of ways to distribute the balls is \(\boxed{1041}\).