2021 AMC 10B Problems/Problem 22
Ang, Ben, and Jasmin each have blocks, colored red, blue, yellow, white, and green; and there are empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives blocks all of the same color is , where and are relatively prime positive integers. What is
Let our denominator be , so we consider all possible distributions.
We use PIE (Principle of Inclusion and Exclusion) to count the successful ones.
When we have at box with all balls the same color in that box, there are ways for the distributions to occur ( for selecting one of the five boxes for a uniform color, for choosing the color for that box, for each of the three people to place their remaining items).
However, we overcounted those distributions where two boxes had uniform color, and there are ways for the distributions to occur ( for selecting two of the five boxes for a uniform color, for choosing the color for those boxes, for each of the three people to place their remaining items).
Again, we need to re-add back in the distributions with three boxes of uniform color... and so on so forth.
Our success by PIE is yielding an answer of .
As In Solution 1, the probability is Dividing by , we get Dividing by , we get Dividing by , we get .
Use complementary counting. Denote as the total number of ways to put colors to boxes by 3 people out of which ways are such that no box has uniform color. Notice . From this setup we see the question is asking for . To find we want to exclude the cases of a) one box of the same color, b) 2 boxes of the same color, c) 3 boxes of same color, d) 4 boxes of the same color, and e) 5 boxes of the same color. Cases d) and e) coincide. From this, we have
In case b), there are ways to choose 2 boxes that have the same color, ways to choose the two colors, 2! ways to permute the 2 chosen colors, and ways so that the remaining 3 boxes don’t have the same color. The same goes for cases a) and c). In case e), the total number of ways to permute 5 colors is . Now, we just need to calculate , and .
We have , since we subtract the number of cases where the boxes contain uniform colors, which is 2.
In the same way, - again, we must subtract the number of ways at least 1 box has uniform color. There are ways if 3 boxes each contains uniform color. Two boxes each contains uniform color is the same as previous. If one box has the same color, there are ways to pick a box, and ways to pick a color for that box, 1! ways to permute the chosen color, and ways for the remaining 2 boxes to have non-uniform colors. Similarly,
Thus, the probability is and the answer is
Video Solution by OmegaLearn (Principle of Inclusion Exclusion)
Video Solution by Interstigation
~ Briefly went over Principal of Inclusion Exclusion using Venn Diagram
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