AMC 10 · 2003 · #21

Grade 7 counting
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
combinations-basicsystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: combinations-basic
📏 Long solution 💡 3 insights
Problem
Pat picks six cookies from a tray stocked with three kinds: chocolate chip, oatmeal, and peanut butter. At least six of each kind are on hand, so no kind can run out. The order they are taken does not matter; only how many of each kind end up in the pile counts. Count how many different assortments of six cookies are possible.

Pick an answer.

(A)
22
(B)
25
(C)
27
(D)
28
(E)
729

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

An assortment is just a choice of three whole numbers (c, o, p) — the counts of chocolate chip, oatmeal, and peanut butter — that add to 6. Tool #2 (Make a Systematic List) is the safe way to count these without missing or double-counting: sweep through every possibility in a fixed order. To keep the list short, Tool #7 (Identify Subproblems) fixes the chocolate chip count c first, which turns the job into a smaller two-kind count for the remaining cookies. Tool #5 (Look for a Pattern) then notices those smaller counts form a tidy run 7, 6, 5, 4, 3, 2, 1, so the grand total is a quick sum instead of a long tally. The tempting 729 = 3⁶ is what you would get if the cookies were taken in order and each of the six picks were independently one of three kinds — but assortments ignore order, so the true count is far smaller.

1STEP 1

Turn cookies into a counting question

Write a pile as three whole numbers: c chocolate chip, o oatmeal, p peanut butter. Six cookies means c + o + p = 6, each count 0 or more.

c + o + p = 6, c, o, p ≥ 0 whole numbers
2STEP 2

Fix the chocolate chip count first

Sort the piles by the chocolate chip count c, which runs 0 through 6. The rest then split two ways: o + p = 6 - c.

For each c ∈ {0,1,2,3,4,5,6}: o + p = 6 - c
3STEP 3

Count the two-kind splits

With leftover L = 6 - c, letting o run 0 to L forces p, so there are L + 1 splits: 7, 6, 5, 4, 3, 2, 1.

#{(o,p): o+p=L} = L+1 → 7,6,5,4,3,2,1
4STEP 4

Add up all the cases

Each pile has exactly one c, so the seven cases are disjoint and complete: 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 assortments, choice (D).

7+6+5+4+3+2+1 = 28 → (D)
Answer
28
The answer 28 sits sensibly among the choices. The trap 729 = 3⁶ would be right only if the six cookies were taken one after another and each pick were freely one of three kinds — but that counts orderings, and here order is ignored, so the real total must be far smaller than 729, ruling out (E). The small distractors 22, 25, 27 are near-misses from undercounting a case or two; the careful case-by-case sweep lands exactly on 28. It also passes a sanity edge check: the all-one-kind piles (all chocolate chip, all oatmeal, all peanut butter) are 3 of the 28, which fits.
💡Key takeaway

When only the counts of each kind matter, list the piles in order by fixing one kind first — here that turns into 7+6+5+4+3+2+1 = 28.

  • Turn cookies into a counting question
  • Fix the chocolate chip count first
  • Count the two-kind splits
  • Add up all the cases

More like this

Same archetype — closest grade level first.