AMC 10 · 2008 · #19

Grade 8 countingalgebra
Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements
generating-functionsfundamental-counting-principlecomplementary-counting complementary-countingsystematic-enumeration ↑ Prerequisites: fundamental-counting-principle
📏 Medium solution 💡 2 insights
Problem
Three brackets of consecutive powers are multiplied out, one long and two short. Find the coefficient of one particular power.

Pick an answer.

(A)
195
(B)
196
(C)
224
(D)
378
(E)
405
How to solve
Strategy Change Focus / Count the Complement

Nobody should expand this product term by term. Expanding means choosing one term from each bracket, and the exponents of the chosen terms add, so the coefficient of x²8 is just a count: how many ways can three legal exponents add to 28. The two short brackets turn out to control everything, because once they are chosen the long bracket's exponent is forced. That turns a three-choice count into a 15 by 15 grid of pairs. Counting the whole grid is easy, so the real work is finding exactly which grid points are illegal and subtracting them. Checking the two boundary walls is what decides the answer, and it is where the problem hides its one surprise.

1STEP 1

Turn the coefficient into a count

A coefficient is really a count of choices.

x^a · x^b · x^c = x^a+b+c, 0 ≤ a ≤ 27, 0 ≤ b ≤ 14, 0 ≤ c ≤ 14
2STEP 2

Let the two short brackets decide

The two short brackets choose freely and the long one is forced.

a + b + c = 28 ⟹ a = 28 - b - c
3STEP 3

Test both walls on the forced a

One end of its range is automatic; the other rules out just one pick.

b+c ≤ 14+14 = 28 ⟹ a = 28-b-c ≥ 0 always; a ≤ 27 ⇔ b+c ≥ 1 ⇔ (b,c) ≠ (0,0)
4STEP 4

Count the whole grid first

The whole grid holds 225 choices.

15 × 15 = 225
5STEP 5

Subtract the one bad corner

Removing the one bad corner gives 224, choice (E).

225 - 1 = 224
Answer
224
Two independent checks. First, shrink the problem to the same shape and test it by hand: use (1+x+x²)²(1+x+x²+x³) and ask for the coefficient of x⁴, where the long bracket again stops one power short of the target. The rule predicts 3² - 1 = 8. Expanding directly, (1+x+x²)² = 1+2x+3x²+2x³+x⁴, and pairing each of its terms of degree 1 through 4 with the matching power in the long bracket gives 2+3+2+1 = 8, matching. Second, cross-check the full problem a different way: (1+x+…+x¹⁴)² has coefficients 1, 2, 3, …, 15, …, 3, 2, 1, whose total is 15² = 225, and the long bracket pairs every one of those degrees except degree 0 with a power of x it actually owns, giving 225 - 1 = 224 again. The trap is worth naming: 225 is the answer a solver gets by forgetting the ceiling at x²⁷, and 225 is deliberately not among the choices, so (C) 224 is the consistent value.
💡Key takeaway

Expanding brackets is just picking one term from each and adding the exponents, so the coefficient is a count: all 225 pairs from the two short brackets work except the single one that would force the long bracket to hand over an x²8 it does not have.

  • Turn the coefficient into a count
  • Let the two short brackets decide
  • Test both walls on the forced a
  • Count the whole grid first
  • Subtract the one bad corner

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