# (a) 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.

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(a) Let's consider the distribution of 21 distinct items into 6 distinct boxes:

i) At least two empty boxes:

Let's use the principle of inclusion and exclusion. We can count the total number of distributions without any restrictions and then subtract the cases where only one box or no box is empty, and add back the cases where two boxes are empty.

Total number of distributions without restrictions: $$6^{21}$$

Number of distributions with only one box empty: $$\binom{6}{1} \cdot 5^{21}$$ (Choose one box to be empty and distribute the items in the remaining 5 boxes)

Number of distributions with no box empty: $$\binom{6}{0} \cdot 6^{21}$$ (All boxes must be filled)

Number of distributions with two boxes empty: $$\binom{6}{2} \cdot 4^{21}$$ (Choose two boxes to be empty and distribute the items in the remaining 4 boxes)

Using the inclusion-exclusion principle:

$\text{Ways with at least two empty boxes} = 6^{21} - \binom{6}{1} \cdot 5^{21} + \binom{6}{0} \cdot 6^{21} - \binom{6}{2} \cdot 4^{21}$

ii) No empty box:

In this case, each box must contain at least one item. This is equivalent to distributing the remaining 15 items (21 - 6) among the 6 boxes.

$\text{Ways with no empty box} = 6^{15}$

So, the total number of ways to distribute 21 distinct items into 6 distinct boxes is given by the sum of the two cases:

$\text{Total ways} = \text{Ways with at least two empty boxes} + \text{Ways with no empty box}$