2010 AMC 12B Problems/Problem 17
- The following problem is from both the 2010 AMC 12B #17 and 2010 AMC 10B #23, so both problems redirect to this page.
Problem
The entries in a array include all the digits from through , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
Solution 1
Observe that all tableaus must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each tableau, there exists a valid tableau diagonally symmetrical across the diagonal extending from the top left to the bottom right.
- Case 1: Center 4
3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases.
- Case 2: Center 5
Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that , logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry,
- Case 3: Center 6
By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured.
~BJHHar
P.S.: I like the tetris approach used in Solution 2 but found it a bit arbitrary. Solution 3 is the best, but not many would know hook length theorem. If the initial observations are unclear, make a tableau with a range of possible numbers in each square.
Solution 2
The first 4 numbers will form one of 3 tetris "shapes".
First, let's look at the numbers that form a block, sometimes called tetris :
Second, let's look at the numbers that form a vertical "L", sometimes called tetris :
Third, let's look at the numbers that form a horizontal "L", sometimes called tetris :
Now, the numbers 6-9 will form similar shapes (rotated by 180 degrees, and anchored in the lower-right corner of the 3x3 grid).
If you match up one tetris shape from the numbers 1-4 and one tetris shape from the numbers 6-9, there is only one place left for the number 5 to be placed.
So what shapes will physically fit in the 3x3 grid, together?
The answer is .
Solution 3
This solution is trivial by the hook length theorem. The hooks look like this:
So, the answer is =
P.S. The hook length formula is a formula to calculate the number of standard Young tableaux of a Young diagram. Numberphile has an easy-to-understand video about it here: https://www.youtube.com/watch?v=vgZhrEs4tuk The full proof is quite complicated and is not given in the video, although the video hints at possible proofs.
Video Solution
https://youtu.be/ZfnxbpdFKjU?t=422
~IceMatrix
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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All AMC 12 Problems and Solutions |
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