# 1997 AIME Problems/Problem 8

## Problem

How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?

## Solution

### Solution 1

For more detailed explanations, see related problem (AIME I 2007, 10).

The problem is asking us for all configurations of $4\times 4$ grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:

• The first two columns share no two numbers in the same row. There are ${4\choose2} = 6$ ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are ${4\choose 2}$ ways. This gives $6^2 = 36$.
• The first two columns share one number in the same row. There are ${4\choose 1} = 4$ ways to pick the position of the shared 1, then ${3\choose 2} = 3$ ways to pick the locations for the next two 1s, and then $2$ ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in $2$ ways. This gives $4 \cdot 3 \cdot 2 \cdot 2 = 48$.
• The first two columns share two numbers in the same row. There are ${4\choose 2} = 6$ ways to pick the position of the shared 1s. Everything is then fixed.

Adding these cases up, we get $36 + 48 + 6 = \boxed{090}$.

### Solution 2

Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of $6$ ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange the one 1 and 2 -1's we have left to put. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is $6*3*(2+1+2)$ = $\boxed{090}$.

-pi_is_3.141

## Solution 3

We can think about it as shading a $4 \times 4$ array so that there are exactly two shaded unit squares in each row and each column. Then the answer is $\boxed{90}$.

## Solution 4

Unlike the above informative solution, this one is hopefully clearer. Notice that for every arrangement $A$ of the first rows of $-1$s and $1$s, we have the inverse of that row $A^{-1}$ so that the sum of the rows and columns of $A$ and $A^{-1}$ is $0$. Therefore if we have another arrangement $B$, we have $B^{-1}$. For instance, if $A=(-1,1,1,-1)$, $A^{-1}=(1,-1,-1,1)$. We then have that if we fix the first row $A$, we have that first there are $\binom{4}{2}$ values of the fixed $A$. We then have the following cases:

Case $1$: ( $AA^{-1}BB^{-1}$). $\binom{4}{2}$.

Case $2$: ( $ABA^{-1}B^{-1}$). $\binom{4}{2}$.

Case $3$: ( $AAA^{-1}A^{-1}$). $\binom{4}{2}/2$, where here we divided by $2$ because then we would overcount $AA$ and $A^{-1}A^{-1}$.

Therefore the answer is $\binom{4}{2}(\binom{4}{2}+\binom{4}{2}+\binom{4}{2}/2)$= $6(6+6+3)$= $\boxed{090}$.

~th1nq3r

• 2007 AIME I Problems/Problem 10 - the same problem, but with a $4\times 6$
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