Difference between revisions of "2007 AIME I Problems/Problem 10"

Revision as of 22:22, 15 March 2007

Problem

In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.

Solution

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Solution 1

Consider the first column. There are ${6\choose3} = 20$ ways that the balls could be chosen, but WLOG let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.

Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases:

All three balls are in the same column. In this case, there are 3 choices for which column that is. From here, the bottom half of the board is fixed.

Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have $3 \cdot 2 \cdot 3 \cdot 3 = 54$ .

All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is not filled in. But there are 6 ways to do this. So in all there are 36 ways.

So there are $20(3+54+36) = 1860$ different shadings, and the solution is $860$ .

Solution 2

There are ${6\choose3}$ to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns:

If column 1 and column 2 do not share any two filled squares on the same row, then there are ${6\choose3}$ combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives ${6\choose3}^2 = 400$ arrangements.
If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row ( $6 - 1 = 5$ places), share another empty square on a row, and have 2 squares each on different rows. This gives $6 \cdot 5 \cdot {4\choose2} = 180$ . Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get ${4\choose2} = 6$ . We get $180 \cdot 6 = 1080$ .
If column 1 and column 2 share 2 filled squares on the same row ( ${6\choose2} = 15$ places), they must also share 2 empty squares on the same row ( ${4\choose2} = 6$ ). The last two squares can be arranged in ${2\choose1} = 2$ positions; this totals to $15 \cdot 6 \cdot 2 = 180$ . Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have ${2\choose1} = 2$ places, giving $180 \cdot 2 = 360$ .
If column 1 and column 2 share 3 filled squares on the same row ( ${6\choose3} = 20$ places), then the squares on columns 3 and 4 are fixed.

Thus, there are $400 + 1080 + 360 + 20 = 1860$ number of shadings, and the solution is $860$ .

Solution 3

We draw a bijection between walking from $(0,0,0,0)$ to $(3,3,3,3)$ as follows: if in the ith row, the jth and kth columns are shaded, then the (2i-1)st step is in the direction corresponding to j, and the (2i)th step is in the direction corresponding to k (j < k) here. We can now use PIE based on the stipulation that j!=k to solve the problem:

$\frac{12!}{3!^42^6}-4\cdot{6 \choose 1}\frac{10!}{3!^32^5}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^22^4}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!2^3}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}\frac{4!}{2^2} = 1860$

So that the answer is $860$ .

@@ Line 31: / Line 31: @@
 We draw a bijection between walking from <math>(0,0,0,0)</math> to <math>(3,3,3,3)</math> as follows: if in the ith row, the jth and kth columns are shaded, then the (2i-1)st step is in the direction corresponding to j, and the (2i)th step is in the direction corresponding to k (j < k) here. We can now use PIE based on the stipulation that j!=k to solve the problem:
-<math>\frac{12!}{3!^42^6}-{4 \choose 1}{6 \choose 1}\frac{10!}{3!^32^5}+{4 \choose 2}{6 \choose 2}\frac{8!}{3!^22^4}-{4 \choose 3}{6 \choose 3}\frac{6!}{3!2^3}+{4 \choose 4}{6 \choose 4}\frac{4!}{2^2} = 1860</math>
+<math>\frac{12!}{3!^42^6}-4\cdot{6 \choose 1}\frac{10!}{3!^32^5}+4\cdot 3\cdot{6 \choose 2}\frac{8!}{3!^22^4}-4\cdot 3\cdot 2\cdot{6 \choose 3}\frac{6!}{3!2^3}+4\cdot 3\cdot 2\cdot 1\cdot {6 \choose 4}\frac{4!}{2^2} = 1860</math>
 So that the answer is <math>860</math>.

2007 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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