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Difference between revisions of "2024 AIME II Problems/Problem 9"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Video Solution==
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https://youtu.be/cCLPjGvhZV4?si=AOsO--AKvONdKrgg
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~MathProblemSolvingSkills.com
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==See also==
 
==See also==

Revision as of 18:38, 23 March 2024

Problem

There is a collection of $25$ indistinguishable white chips and $25$ indistinguishable black chips. Find the number of ways to place some of these chips in the $25$ unit cells of a $5\times5$ grid such that:

  • each cell contains at most one chip
  • all chips in the same row and all chips in the same column have the same colour
  • any additional chip placed on the grid would violate one or more of the previous two conditions.

Solution 1

The problem says "some", so not all cells must be occupied. We start by doing casework on the column on the left. There can be 5,4,3,2, or 1 black chip. The same goes for white chips, so we will multiply by 2 at the end. There is $1$ way to select $5$ cells with black chips. Because of the 2nd condition, there can be no white, and the grid must be all black- $1$ way . There are $5$ ways to select 4 cells with black chips. We now consider the row that does not contain a black chip. The first cell must be blank, and the remaining $4$ cells have $2^4-1$ different ways($-1$ comes from all blank). This gives us $75$ ways. Notice that for 3,2 or 1 black chips on the left there is a pattern. Once the first blank row is chosen, the rest of the blank rows must be ordered similarly. For example, with 2 black chips on the left, there will be 3 blank rows. There are 15 ways for the first row to be chosen, and the following 2 rows must have the same order. Thus, The number of ways for 3,2,and 1 black chips is $10*15$, $10*15$, $5*15$. Adding these up, we have $1+75+150+150+75 = 451$. Multiplying this by 2, we get $\boxed{902}$. ~westwoodmonster

Solution 2

Note that the answer is equivalent to the number of ways to choose rows and columns that the white chips occupy, as once those are chosen, there is only one way to place the chips, and every way to place the chips corresponds to a set of rows and columns occupied by the white pieces.

If the white pieces occupy none of the rows, then because they don't appear on the board, they will not occupy any of the columns. Similar logic can be applied to show that if white pieces occupy all of the rows, they will also occupy all of the columns.

The number of sets of rows and columns that white can occupy are $2^{5} - 2 = 30$ each, accounting for the empty and full set. So, including the board with 25 white pieces and the board with 25 black pieces, the answer is $30^{2}+2 = \boxed{902}$

Solution 3

Case 1: All chips on the grid have the same color.

In this case, all cells are occupied with chips with the same color. Therefore, the number of configurations in this case is 2.

Case 2: Both black and white chips are on the grid.

Observation 1: Each colored chips must occupy at least one column and one row.

This is because, for each given color, there must be at least one chip. Therefore, all chips placed in the cells that are in the same row or the same column with this given chip must have the same color with this chip.

Observation 2: Each colored chips occupy at most 4 rows and 4 columns.

This directly follows from Observation 1.

Observation 3: For each color, if all chips in this color occupy columns with $x$-coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$-coordinates $\left\{ y_1, \cdots, y_n \right\}$, then every cell $\left( x, y \right)$ with $x \in \left\{ x_1, \cdots , x_m \right\}$ and $y \in \left\{ y_1, \cdots , y_n \right\}$ is occupied by a chip with the same color.

This is because, if there is any cell in this region occupied by a chip with a different color, it violates Condition 2. If there is any cell in this region that is empty, then it violates Condition 3.

Observation 4: For each color, if all chips in this color occupy columns with $x$-coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$-coordinates $\left\{ y_1, \cdots, y_n \right\}$, then every cell $\left( x, y \right)$ with $x \notin \left\{ x_1, \cdots , x_m \right\}$ and $y \in \left\{ y_1, \cdots , y_n \right\}$, or $x \in \left\{ x_1, \cdots , x_m \right\}$ and $y \notin \left\{ y_1, \cdots , y_n \right\}$ is empty.

This is because, if there is any cell in this region occupied by a chip with a different color, it violates Condition 2.

Observation 5: For each color, if all chips in this color occupy columns with $x$-coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$-coordinates $\left\{ y_1, \cdots, y_n \right\}$, then every cell $\left( x, y \right)$ with $x \notin \left\{ x_1, \cdots , x_m \right\}$ and $y \notin \left\{ y_1, \cdots , y_n \right\}$ is occupied by chips with the different color.

This follows from Condition 3.

By using the above observations, the number of feasible configurations in this case is given by \begin{align*} \sum_{n=1}^4 \sum_{m=1}^4 \binom{5}{n} \binom{5}{m} & = \left( \sum_{n=1}^4 \binom{5}{n} \right) \left( \sum_{m=1}^4 \binom{5}{m} \right) \\ & = \left( 2^5 - 2 \right)^2 \\ & = 900 . \end{align*}

Putting all cases together, the total number of feasible configurations is $2 + 900 = \boxed{\textbf{(902) }}$.


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/6gTAyjbCLh4

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)


Video Solution

https://youtu.be/cCLPjGvhZV4?si=AOsO--AKvONdKrgg

~MathProblemSolvingSkills.com


See also

2024 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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