Difference between revisions of "2010 AMC 12B Problems/Problem 17"
Mathpro12345 (talk | contribs) (→Solution 1) |
m |
||
Line 66: | Line 66: | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=16|num-a=18|ab=B}} | {{AMC12 box|year=2010|num-b=16|num-a=18|ab=B}} | ||
+ | {{AMC10 box|year=2010|num-b=22|num-a=24|ab=B}} | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:37, 22 December 2020
- 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
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 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.