practice problems pending
1. How many ways to distribute 10 different balls into 3 different bags such that none of the bags is empty? S1. Distinct objects/distinct buckets => PIE(Principle of Inclusion Exclusion) Step 1: Total cases: 3^10 since each ball can go into any of the bags. Step 2: Subtract those cases where one bag is empty: 3C1 to choose one bag. And now each ball has 2 options: 2^10. So total 3*2^10 cases need to be subtracted to account for one empty bucket. In step 2 consider what happens in detail: When we made one bucket empty, 10 balls have to go into 2 buckets: 2^10. This also includes the cases when all the balls go to one bucket. So considering empty bucket1 also considers the cases where bucket 1 and 2 are both empty. Similarly when we consider bucket2 as empty we are also considering the case where both bucket 1 and 2 are empty. Essentially we are double subtracting the cases where 2 buckets are empty. So we need to add them back. 3C2 to choose 2 buckets to remain empty. Now each ball ...