How many ways are there to distribute 21 district items into 6 distinct boxes with: 
i) At least two empty box.
ii) No empty box.

i) Distributing 21 items into 6 distinct boxes with at least two empty boxes:

To solve this problem, we can use a combinatorial approach. We'll consider the cases where exactly 2, 3, 4, 5, or 6 boxes are empty, and then subtract these cases from the total possibilities.

Case 1: Exactly 2 boxes are empty: Choose 2 out of 6 boxes to be empty: ${6 \choose 2} = 15$ ways. For the remaining 4 boxes, we distribute 21 items among them, which is equivalent to __________ __________ __________ ______ ___ _______ ____ ______.
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