Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 12B Problems/Problem 17"

(→‎Solution 3 (Doesn't require the hook length formula, similar to Solution 1))
(→‎Solution 3 (Doesn't require the hook length formula, copied from Solution 1 and the Beauty of Math video))
 
(17 intermediate revisions by the same user not shown)
Line 69: Line 69:
 
~edits by glowworm
 
~edits by glowworm
  
−
== Solution 3 (Doesn't require the hook length formula, copied from Solution 1)==
+
== Solution 3 (Doesn't require the hook length formula, copied from Solution 1 and the Beauty of Math video)==
  
 
As Solution number one and the videos said 1 and 9 must go on the top left and top right.
 
As Solution number one and the videos said 1 and 9 must go on the top left and top right.
 
The middle must be 4, 5, or 6 because there must be two numbers less than it to fill the top middle and left middle squares (so 2 and 3 can't be in the middle) and two numbers more than it to fil the right middle and bottom middle squares (7 and 8 don’t work).
 
The middle must be 4, 5, or 6 because there must be two numbers less than it to fill the top middle and left middle squares (so 2 and 3 can't be in the middle) and two numbers more than it to fil the right middle and bottom middle squares (7 and 8 don’t work).
  
−
So, first casework with the middle being 4.
+
So, first casework with the middle being 4. 2 and 3 must go in the middle top and the middle left boxes because they are the only numbers smaller than 4. There are 2 ways to arrange the 2 and 3 because they can flip their places in the middle top and middle left boxes.  
−
<math></math>\begin{tabular}{|c|c|c|} \hline 1&&\\
+
<cmath>\begin{tabular}{|c|c|c|}  
−
\hline &4&\\
+
\hline 1&2&\\
 +
\hline 3&4&\\
 
\hline &&9\\
 
\hline &&9\\
 +
\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&3&\\
 +
\hline 2&4&\\
 +
\hline &&9\\
 +
\hline \end{tabular}</cmath>
 +
Now, for the rightmost column, choose 2 numbers from the 4 remaining numbers (5, 6, 7, and 8) to fill in that column. Since those two numbers have to increase going downwards, there is only one way to arrange those 2 numbers. The two remaining numbers must automatically go to the two spots remaining on the bottom row, and likewise there is only one way to arrange them. For example, if I picked 5 and 8 to go in the 2 spots in the rightmost square, I would have one of these two charts (there were two ways to arrange the 2 and 3 in the chart):
 +
<cmath>\begin{tabular}{|c|c|c|}
 +
\hline 1&2&5\\
 +
\hline 3&4&8\\
 +
\hline 6&7&9\\
 +
\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&3&5\\
 +
\hline 2&4&8\\
 +
\hline 6&7&9\\
 +
\hline \end{tabular}</cmath>
 +
So, the number of ways to make a chart where the middle number is 4 is <math>2 \cdot \binom{4}{2} = 2\cdot6 = 12</math> ways.
 +
 +
For the case when the middle is 6, it is exactly symmetric to the middle being 4 but on the opposite side. 7 and 8 must be in the bottom middle and right middle squares because if you tried to put any other number in those two squares that number would have to be bigger than 7 and 8 but smaller than 9. So there are 2 ways to arrange 7 and 8 (because they can flip themselves) and 4 choose 2 ways to choose two numbers (out of 2, 3, 4, and 5) to go in the two spaces available on the leftmost column. Again, the remaining two numbers will be forced to go in only one way on the two spaces available on the top row, meaning there are also <math>12</math> ways to make a chart with the middle number being 6. Example- 7 and 8 are in the bottom middle and right middle squares respectively and I chose 2 and 3 to be the numbers on the two spots on the leftmost column:
 +
<cmath>\begin{tabular}{|c|c|c|}
 +
\hline 1&4&5\\
 +
\hline 2&6&8\\
 +
\hline 3&7&9\\
 +
\hline \end{tabular}</cmath>
  
 +
The case when the middle is 5 is the hardest case.
 +
The numbers 2, 3, and 4 will fill in spaces from the top left until the diagonal with 5, and either the top right or bottom left square will be left blank (no blanks before that because the box must increase). So there are 2 choices which one to leave blank:
 +
<cmath>\begin{tabular}{|c|c|c|}
 +
\hline 1&&\\
 +
\hline &5&\\
 +
\hline x&&9\\
 +
\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&&x\\
 +
\hline &5&\\
 +
\hline &&9\\
 +
\hline \end{tabular}</cmath>
 +
There are 3 choices for the box (either 2, 3, or 4) immediately next to the blank box between 1 and the blank box and the other 2 numbers (from 2, 3, or 4) will sort themselves. For example, if I chose the blank box to be the bottom left box and I chose the number immediately next to the blank box to be 2, the grid would look like this:
 +
<cmath>\begin{tabular}{|c|c|c|}
 +
\hline 1&3&4\\
 +
\hline 2&5&\\
 +
\hline &&9\\
 +
\hline \end{tabular}</cmath>
 +
Now there are 2 empty spaces in either the rightmost or the bottommost row (the row with the square you chose to be blank and 9)  and 1 empty space in the middle of the other row with 9 and a value of 2, 3, or 4. Example- the two empty spaces are marked with an p and the one empty space in the other row is marked with a q:
 +
<cmath>\begin{tabular}{|c|c|c|}
 +
\hline 1&3&4\\
 +
\hline 2&5&q\\
 +
\hline p&p&9\\
 +
\hline \end{tabular}</cmath>
 +
Finally, pick either 6, 7, or 8 to fill the box marked q (the one empty space between a value of 2, 3, or 4) and the remaining two numbers from 6, 7 or 8 will sort itself in the two boxes marked with a p. This is 3 choose 1 = 3 ways. Thus, there are 2 ways to choose the blank box, 3 ways to choose the number (2, 3, or 4) between the blank box and 1, and 3 ways to choose the box on the opposite side of the number between the blank box and 1, and all the other numbers are forced to arrange themselves in 1 way. This means there are <math>2\cdot3\cdot3=18</math> ways to make a chart with 5 in the middle.
  
 +
The answer to the problem is <math>12+12+18 = \boxed{\text{(D) }42}</math> ways to make a chart in total.
  
−
For middle is 5, the numbers 2, 3, and 4 will fill in spaces until the diagonal with 5, and either the top right or bottom left square will be left blank (no blanks before that because the box must increase). So there are 2 choices which one to leave blank. There are 3 choices for the box (either 2, 3, or 4) immediately next to the blank box and the other 2 numbers (from 2, 3, or 4) will sort themselves. Now there are 2 empty spaces in one horizontal or vertical row with 9, and 1 empty space in the middle of the other row with 9 and a value of 2, 3, or 4. Pick either 6, 7, or 8 to fil that box and the other row will sort itself with the remaining two numbers. Thus, there are 2x3x3=18 ways.
+
~unhappyfarmer
−
For middle is 4, the numbers 2 and 3 must go in the middle top and the middle left boxes because those are the only boxes smaller than 4. For the rightmost row, choose 2 from 4 to fill in that row, and on the bottommost row the other two numbers will sort themselves. So there are 4C2 = 6. 6x2 = 12
 
−
For middle is 6, it is symmetric to the middle being 4 because 7 or 8 must be in the middle bottom or middle right. Then 4C2 for one of the symmetric other part of the row with 9, and then the numbers wil sort themselves. 6x2 = 12
 
−
18+12+12=42
 
  
 
== Video Solution by Pi Academy (Fast and Easy) ==
 
== Video Solution by Pi Academy (Fast and Easy) ==

Latest revision as of 20:17, 12 October 2024

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 $3 \times 3$ array include all the digits from $1$ through $9$, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$

Solution 1

Observe that all tables 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 table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right.


  • Case 1: Center 4

\[\begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&\\ \hline &8&9\\ \hline \end{tabular}\]

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. $2*6=12$

  • Case 2: Center 5

\[\begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&8\\ \hline &&9\\ \hline \end{tabular}\]

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 $4<5$, logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2*9=18$

  • Case 3: Center 6

By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2*6=12$

\[12+18+12=\boxed{\textbf{(D) }42}\]


~BJHHar

Solution 2

This solution is trivial by the hook length theorem. The hooks look like this:

$\begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\ \hline 4 & 3 & 2\\ \hline 3 & 2 & 1\\ \hline \end{tabular}$

So, the answer is $\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}$ = $\boxed{\text{(D) }42}$

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.

Hook length theorem: take any shape made out of congruent squares and say the rules are just like the problem described. Now if you count how many squares are to the right and to the down of it, INCLUDING THE NUMBER ITSELF, then multiply the numbers that have been written down, that is the denominator of the fraction. The numerator is simpler: the factorial of the number of squares. Ex:

$\begin{tabular}{|c|c|c|} \hline 4 & 3 & 2\\ \hline 3 & 2 & 1\\ \hline \end{tabular}$ Therefore answer will be 6!/(4 * 3 * 3 * 2 * 2)

~edits by glowworm

Solution 3 (Doesn't require the hook length formula, copied from Solution 1 and the Beauty of Math video)

As Solution number one and the videos said 1 and 9 must go on the top left and top right. The middle must be 4, 5, or 6 because there must be two numbers less than it to fill the top middle and left middle squares (so 2 and 3 can't be in the middle) and two numbers more than it to fil the right middle and bottom middle squares (7 and 8 don’t work).

So, first casework with the middle being 4. 2 and 3 must go in the middle top and the middle left boxes because they are the only numbers smaller than 4. There are 2 ways to arrange the 2 and 3 because they can flip their places in the middle top and middle left boxes. \[\begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&3&\\ \hline 2&4&\\ \hline &&9\\ \hline \end{tabular}\] Now, for the rightmost column, choose 2 numbers from the 4 remaining numbers (5, 6, 7, and 8) to fill in that column. Since those two numbers have to increase going downwards, there is only one way to arrange those 2 numbers. The two remaining numbers must automatically go to the two spots remaining on the bottom row, and likewise there is only one way to arrange them. For example, if I picked 5 and 8 to go in the 2 spots in the rightmost square, I would have one of these two charts (there were two ways to arrange the 2 and 3 in the chart): \[\begin{tabular}{|c|c|c|} \hline 1&2&5\\ \hline 3&4&8\\ \hline 6&7&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&3&5\\ \hline 2&4&8\\ \hline 6&7&9\\ \hline \end{tabular}\] So, the number of ways to make a chart where the middle number is 4 is $2 \cdot \binom{4}{2} = 2\cdot6 = 12$ ways.

For the case when the middle is 6, it is exactly symmetric to the middle being 4 but on the opposite side. 7 and 8 must be in the bottom middle and right middle squares because if you tried to put any other number in those two squares that number would have to be bigger than 7 and 8 but smaller than 9. So there are 2 ways to arrange 7 and 8 (because they can flip themselves) and 4 choose 2 ways to choose two numbers (out of 2, 3, 4, and 5) to go in the two spaces available on the leftmost column. Again, the remaining two numbers will be forced to go in only one way on the two spaces available on the top row, meaning there are also $12$ ways to make a chart with the middle number being 6. Example- 7 and 8 are in the bottom middle and right middle squares respectively and I chose 2 and 3 to be the numbers on the two spots on the leftmost column: \[\begin{tabular}{|c|c|c|} \hline 1&4&5\\ \hline 2&6&8\\ \hline 3&7&9\\ \hline \end{tabular}\]

The case when the middle is 5 is the hardest case. The numbers 2, 3, and 4 will fill in spaces from the top left until the diagonal with 5, and either the top right or bottom left square will be left blank (no blanks before that because the box must increase). So there are 2 choices which one to leave blank: \[\begin{tabular}{|c|c|c|} \hline 1&&\\ \hline &5&\\ \hline x&&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&&x\\ \hline &5&\\ \hline &&9\\ \hline \end{tabular}\] There are 3 choices for the box (either 2, 3, or 4) immediately next to the blank box between 1 and the blank box and the other 2 numbers (from 2, 3, or 4) will sort themselves. For example, if I chose the blank box to be the bottom left box and I chose the number immediately next to the blank box to be 2, the grid would look like this: \[\begin{tabular}{|c|c|c|} \hline 1&3&4\\ \hline 2&5&\\ \hline &&9\\ \hline \end{tabular}\] Now there are 2 empty spaces in either the rightmost or the bottommost row (the row with the square you chose to be blank and 9) and 1 empty space in the middle of the other row with 9 and a value of 2, 3, or 4. Example- the two empty spaces are marked with an p and the one empty space in the other row is marked with a q: \[\begin{tabular}{|c|c|c|} \hline 1&3&4\\ \hline 2&5&q\\ \hline p&p&9\\ \hline \end{tabular}\] Finally, pick either 6, 7, or 8 to fill the box marked q (the one empty space between a value of 2, 3, or 4) and the remaining two numbers from 6, 7 or 8 will sort itself in the two boxes marked with a p. This is 3 choose 1 = 3 ways. Thus, there are 2 ways to choose the blank box, 3 ways to choose the number (2, 3, or 4) between the blank box and 1, and 3 ways to choose the box on the opposite side of the number between the blank box and 1, and all the other numbers are forced to arrange themselves in 1 way. This means there are $2\cdot3\cdot3=18$ ways to make a chart with 5 in the middle.

The answer to the problem is $12+12+18 = \boxed{\text{(D) }42}$ ways to make a chart in total.

~unhappyfarmer

Video Solution by Pi Academy (Fast and Easy)

https://youtu.be/3gwxQ1fjxQM?si=oL0HxnoYSgyl1Rh6

~ Pi Academy

Video Solution 2

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_17&oldid=229711"