AMC 10B 2021: Principle of Inclusion-Exclusion

Question

Problem statement:

Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ 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 $3$ blocks all of the same color is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$

Solution:

Let our denominator be $(5!)^3$, so we consider all possible distributions.

We use PIE (Principle of Inclusion and Exclusion) to count the successful ones.

When we have at 1 box with all 3 balls the same color in that box, there are $\binom{5}{1} \cdot {}_5P_1 \cdot (4!)^3$ ways for the distributions to occur ($\binom{5}{1}$ for selecting one of the five boxes for a uniform color, ${}_5P_1$ for choosing the color for that box, $4!$ 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 $\binom{5}{2} \cdot {}_5P_2 \cdot (3!)^3$ ways for the distributions to occur ($\binom{5}{2}$ for selecting two of the five boxes for a uniform color, ${}_5P_2$ for choosing the color for those boxes, $3!$ 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

$$\binom{5}{1} \cdot {}_5P_1 \cdot (4!)^3 - \binom{5}{2} \cdot {}_5P_2 \cdot (3!)^3 + \binom{5}{3} \cdot {}_5P_3 \cdot (2!)^3 - \binom{5}{4} \cdot {}_5P_4 \cdot (1!)^3 + \binom{5}{5} \cdot {}_5P_5 \cdot (0!)^3 = 120 \cdot 2556.$$

$$\frac{120 \cdot 2556}{120^3} = \frac{71}{400},$$ yielding an answer of D (471).

My question: surely if you have four boxes which contain 3 blocks of the same colour, that automatically implies that the fifth box must have 3 blocks of the same colour too. This is because there are 15 blocks in total (3 for each colour). But in the calculation, $\binom{5}{4} \cdot {}_5P_4 \cdot (1!)^3 \neq \binom{5}{5} \cdot {}_5P_5 \cdot (0!)^3$

I am curious about why there is a discrepancy here.

Source for problem and solution: https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22?srsltid=AfmBOorAuL5KPJM5VgzNOyp2AsuXUrD3fZ3tuhB30YzG6qYPQD2Iz7YW

Answer 1

You have a basic misunderstanding of the justification for the Inclusion-Exclusion formula.

Before you continue reading this response, I suggest that you first skim this article for an introduction to Inclusion-Exclusion, and then study closely this answer.

The second link provides an explanation of and justification for the Inclusion-Exclusion formula.

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The point that you made, that if 4 boxes have all matching colors, then the 5th box must also match is true but irrelevant.

If you study the 2nd linked answer closely, you will see that (in the parlance of the answer), with respect to this particular math problem, $~T_4~$ must be greater than $~T_5.~$

That is, the justification of the formula is that (again in the parlance of the linked article) it forces each individual element in $~| ~S_1 \cup S_2 \cup S_3 \cup S_4 \cup S_5 ~| ~$ to end up being counted exactly once (after all the adding and subtracting).

By the way, I suggest that you write a computer program to sanity check what the exact answer actually is.

I also suggest that you select a specific element that is in (for example) $~S_1 \cap S_2 \cap S_3 \cap S_4~$ and manually check how many times that element is actually counted (after all of the adding and subtracting). Note that, as you indicated, the exact same element will also be in (for example) $~S_2 \cap S_3 \cap S_4 \cap S_5.$