2021 AMC 10B Problems/Problem 22

Problem

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 $m + n ?$

$\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542$

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 $_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3$ ways for the distributions to occur ( $_{5} C _{1}$ for selecting one of the five boxes for a uniform color, $_{5} P _{1}$ 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 $_{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3$ ways for the distributions to occur ( $_{5} C _{2}$ for selecting two of the five boxes for a uniform color, $_{5} P _{2}$ 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 $_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3 - _{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3 + _{5} C _{3} \cdot _{5} P _{3} \cdot (2!)^3 - _{5} C _{4} \cdot _{5} P _{4} \cdot (1!)^3 + _{5} C _{5} \cdot _{5} P _{5} \cdot (0!)^3 = 120 \cdot 2556.$ $\frac{120 \cdot 2556}{120^3}=\frac{71}{400},$ yielding an answer of $\boxed{471 \textbf{(D)}}$ .

Solution 2

As In Solution 1, the probability is $\frac{\binom{5}{1}\cdot 5\cdot (4!)^3 - \binom{5}{2}\cdot 5\cdot 4\cdot (3!)^3 + \binom{5}{3}\cdot 5\cdot 4\cdot 3\cdot (2!)^3 - \binom{5}{4}\cdot 5\cdot 4\cdot 3\cdot 2\cdot (1!)^3 + \binom{5}{5}\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{(5!)^3}$ $= \frac{5\cdot 5\cdot (4!)^3 - 10\cdot 5\cdot 4\cdot (3!)^3 + 10\cdot 5\cdot 4\cdot 3\cdot (2!)^3 - 5\cdot 5! + 5!}{(5!)^3}.$ Dividing by $5!$ , we get $\frac{5\cdot (4!)^2 - 10\cdot (3!)^2 + 10\cdot (2!)^2 - 5 + 1}{(5!)^2}.$ Dividing by $4$ , we get $\frac{5\cdot 6\cdot 24 - 10\cdot 9 + 10 - 1}{30\cdot 120}.$ Dividing by $9$ , we get $\frac{5\cdot 2\cdot 8 - 10 + 1}{10\cdot 40} = \frac{71}{400} \implies \boxed{\textbf{(D) }471}$ .

Solution 3

Use complimentary counting. Denote $T_n$ as the total number of ways to put n colors to n boxes by 3 people out of which $f_n$ ways are such that no box has uniform color. Notice $T_n = (n!)^3$ . From this setup we see the question is asking for $1-\frac{f_5}{(5!)^3}$ . To find $f_5$ 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 e) 5 boxes of the same color. Cases d) and e) coincide. From this, we have

$f_5=(5!)^3-{\binom{5}{1}\binom{5}{1} f_4 - \binom{5}{2}\binom{5}{2}2!f_3 - \binom{5}{3}\binom{5}{3}3!f_2 - 5!}$

In case b), there are $\binom{5}{2}$ ways to choose 2 boxes that will have the same color, $\binom{5}{2}$ ways to choose that color, 2! ways to permute the 2 chosen colors, and $f_3$ ways so that the remaining boxes don’t have the same color. The same goes for cases a) and c). In case e), the total number of ways is $5!$ .

Going back to $f_5$ , there are exactly $(5!)^3$ ways for 3 people to rearrange the 5 colors, but we wish to subtract the cases where at least 1 box has the same color. Now, we just need to calculate $f_2, f_3 and f_4$ .

$f_2=T_2-2 = (2!)^3 - 2 = 6$ , since we subtract the number of cases where the boxes contain uniform colors, which is 2.

$f_3=T_3-[3!+ \binom{3}{1}\binom{3}{1}\cdot f_2] = 216 - [6 + 9\cdot 6] = 156$ - again, we must subtract the ways where boxes have uniform colors. If all 3 boxes have the same color, there are $3!$ ways to permute them. If one box has the same color, there are $\binom{3}{1}$ ways to pick a box, and $\binom{3}{1}$ ways to pick a color for that box. There is 1! ways to rearrange the color in the box. The remaining boxes must have different colors, so we multiply by $f_2$

$f_4=(4!)^3-[4!+ \binom{4}{2}\binom{4}{2}\cdot 2! \cdot f_2+ \binom{4}{1}\binom{4}{1}\cdot f_3] = 13824 - [24 + 36\cdot 2 \cdot 6 + 16 \cdot 156] = 10,872$

Thus, $f_5 = f_5=(5!)^3-{\binom{5}{1}\binom{5}{1}\cdot f_4+ \binom{5}{2}\binom{5}{2}\cdot 2!\cdot f_3+\binom{5}{3}\binom{5}{3}\cdot 3!\cdot f_2 + 5!} = (5!)^3 - [25\cdot 10,872 + 100\cdot 2\cdot 156 + 100\cdot 6\cdot 6 + 120] = (5!)^3 - 306,720$

Thus, the probability is $1 - \frac{(5!)^3 - 306,720}{(5!)^3} = 71/400 and the answer is \boxed{\textbf{(D) }471}$

Video Solution by OmegaLearn (Principal of Inclusion Exclusion)

https://youtu.be/o0S8SqRO0Yc

~ pi_is_3.14

Video Solution by Interstigation

https://youtu.be/OVW9KhmmrVQ

~ Briefly went over Principal of Inclusion Exclusion using Venn Diagram

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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