AMC 10 · 2010 · #23

Grade 7 counting
systematic-enumerationcombinations-basicsymmetry-argument casework ↑ Prerequisites: systematic-enumeration
📏 Long solution 💡 3 insights
Problem
Fill a 3 × 3 grid with the digits 1 through 9, each used once, so that every row reads left-to-right in increasing order and every column reads top-to-bottom in increasing order. Count how many different grids are possible.

Pick an answer.

(A)
18
(B)
24
(C)
36
(D)
42
(E)
60

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

"How many arrangements" is the classic trigger for Tool #2 (Make a Systematic List). Nine digits in nine cells is too many to list blindly, so first Tool #14 (Extreme Principle) pins the forced cells: the top-left must be the smallest digit and the bottom-right the largest, and the center is squeezed to just three possible values. That splits the count into a few clean cases — Tool #7 (Identify Subproblems) — organized by the center digit. Inside each case the remaining digits fall into small ordered groups that are easy to count with the multiplication principle. Tool #5 (Look for a Pattern) then reuses one case for its mirror image, halving the work.

1STEP 1

Pin the two corners

Cell a is smaller than all eight other digits, so a is 1; cell i is larger than all eight, so i is 9.

a = 1, i = 9
2STEP 2

Trap the center

Three cells (a, b, d) must be below the center e and three (f, h, i) above it, so e is 4, 5, or 6.

a,b,d < e < f,h,i → 4 ≤ e ≤ 6
3STEP 3

Split by the center digit

Count the grids with center 4, with center 5, and with center 6 separately, then add the three totals.

Total = N(e=4) + N(e=5) + N(e=6)
4STEP 4

Count center = 4

Cells b and d take 2 and 3 in either order (2 ways), and picking which two of 5, 6, 7, 8 go in the right column fixes the rest: 12.

N(e=4) = 2 × C(4, 2) = 2 × 6 = 12
5STEP 5

Count center = 6 by symmetry

Swapping each digit k for 10-k and flipping the grid pairs every center-4 grid with exactly one center-6 grid, so this case also gives 12.

k ↦ 10-k → N(e=6) = N(e=4) = 12
6STEP 6

Count center = 5

b and d are small, f and h are large, and one of c, g takes the last small digit; each choice allows 3 × 3 = 9, so 18.

N(e=5) = (3 × 3) + (3 × 3) = 9 + 9 = 18
7STEP 7

Add the cases

Add the three case counts, 12 + 18 + 12, for the total number of valid grids: 42.

12 + 18 + 12 = 42 → (D)
Answer
42
The three cases e=4,5,6 are exhaustive (the center can be nothing else) and mutually exclusive (a grid has one center digit), so adding 12+18+12=42 is legitimate. The symmetry between e=4 and e=6 shows up as equal counts, a good internal consistency check. The total 42 matches choice (D), and it also agrees with the standard result that the number of increasing 3×3 fillings (standard Young tableaux of a 3×3 square) is 9!/(5·4·3·4·3·2·3·2·1) = 362880/8640 = 42.
💡Key takeaway

Lock the forced cells first — 1 in the top-left, 9 in the bottom-right, and the center can only be 4, 5, or 6 — then count each center case separately: 12+18+12=42.

  • Pin the two corners
  • Trap the center
  • Split by the center digit
  • Count center = 4
  • Count center = 6 by symmetry
  • Count center = 5
  • Add the cases