AMC 10 · 2003 · #21
Grade 7 countingPat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Pat picks six cookies from a tray that has three kinds available — chocolate chip, oatmeal, and peanut butter — with at least six of every kind on hand. Only how many of each kind end up in the pile matters, not the order they are taken. Count how many different piles of six are possible.
Givens: Exactly six cookies are chosen in total.; There are three kinds to choose from: chocolate chip, oatmeal, peanut butter.; At least six of each kind are available, so no kind can run out.; An assortment is described only by how many of each kind it contains, so order does not matter.; Answer choices: (A) $22$, (B) $25$, (C) $27$, (D) $28$, (E) $729$.
Unknowns: The number of different assortments of six cookies, i.e. the number of ways to say how many chocolate chip, oatmeal, and peanut butter cookies are in the pile.
Understand
Restated: Pat picks six cookies from a tray that has three kinds available — chocolate chip, oatmeal, and peanut butter — with at least six of every kind on hand. Only how many of each kind end up in the pile matters, not the order they are taken. Count how many different piles of six are possible.
Givens: Exactly six cookies are chosen in total.; There are three kinds to choose from: chocolate chip, oatmeal, peanut butter.; At least six of each kind are available, so no kind can run out.; An assortment is described only by how many of each kind it contains, so order does not matter.; Answer choices: (A) $22$, (B) $25$, (C) $27$, (D) $28$, (E) $729$.
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #5 Look for a Pattern
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 $729 = 3^6$ 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.
Execute — Answer: D
7.SP.C.8 Step 1 Turn cookies into a counting question
- Describe any pile by three whole numbers: $c$ chocolate chip, $o$ oatmeal, $p$ peanut butter.
- Since the pile has six cookies, these must satisfy $c + o + p = 6$, and each is $0$ or more.
- Two piles are the same assortment exactly when they have the same three counts, so counting assortments means counting the whole-number solutions of $c + o + p = 6$.
💡 A pile of cookies is fully pinned down by how many of each kind it holds, so listing piles is the same as listing number triples that add to six.
7.SP.C.8 Step 2 Fix the chocolate chip count first
- Sort all the piles by how many chocolate chip cookies they contain.
- That count $c$ can be $0, 1, 2, 3, 4, 5,$ or $6$.
- Once $c$ is chosen, the remaining $6 - c$ cookies must be split between just two kinds, oatmeal and peanut butter, with $o + p = 6 - c$.
- So the big three-kind problem breaks into seven smaller two-kind problems, one for each value of $c$.
💡 Locking one kind's count first shrinks a hard three-way split into an easy two-way split you already know how to count.
4.OA.C.5 Step 3 Count the two-kind splits
- For a fixed leftover $L = 6 - c$, how many ways can $o + p = L$ with whole numbers?
- Let $o$ run from $0$ up to $L$; each choice of $o$ forces $p = L - o$.
- That gives $L + 1$ splits.
- So $c = 0$ leaves $L = 6$ and gives $7$ piles; $c = 1$ gives $6$; then $5, 4, 3, 2$; and $c = 6$ leaves $L = 0$ and gives just $1$ pile (all chocolate chip).
- The counts are $7, 6, 5, 4, 3, 2, 1$.
💡 Splitting $L$ cookies into two named piles just means picking how many go in the first pile — anything from $0$ to $L$, which is $L+1$ choices.
4.OA.A.3 Step 4 Add up all the cases
- The seven cases cover every assortment once and never overlap, because each pile has exactly one chocolate chip count.
- Add the counts: $7 + 6 + 5 + 4 + 3 + 2 + 1 = 28$.
- So there are $28$ different assortments of six cookies, which is choice (D).
💡 Because the seven cases are separate and complete, the total is simply their sum.
7.SP.C.8 Describe any pile by three whole numbers: $c$ chocolate chip, $o$ oatmeal, $p$ p 7.SP.C.8 Sort all the piles by how many chocolate chip cookies they contain. That count $ 4.OA.C.5 For a fixed leftover $L = 6 - c$, how many ways can $o + p = L$ with whole numbe 4.OA.A.3 The seven cases cover every assortment once and never overlap, because each pile Review
Reasonableness: The answer $28$ sits sensibly among the choices. The trap $729 = 3^6$ 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.
Alternative: Stars and bars: picture the six cookies as six stars in a row and drop in two dividers to mark where chocolate chip ends and peanut butter begins. Every arrangement of six stars and two dividers gives one assortment, and there are $\binom{6+2}{2} = \binom{8}{2} = 28$ ways to place the two dividers among the eight slots — the same $28$, found in one step.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Turning the cookie choice into an organized count of the whole-number triples $(c,o,p)$ that sum to $6$, swept case by case so nothing is missed or repeated.)4.OA.C.5Generate a number or shape pattern following a given rule (Seeing that fixing the chocolate chip count gives $L+1$ two-kind splits, so the case counts follow the pattern $7, 6, 5, 4, 3, 2, 1$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Adding the seven separate case counts $7+6+5+4+3+2+1$ to reach the total $28$.)
⭐ When only the counts of each kind matter, list the piles in order by fixing one kind first — here that turns into $7+6+5+4+3+2+1 = 28$.
⭐ When only the counts of each kind matter, list the piles in order by fixing one kind first — here that turns into $7+6+5+4+3+2+1 = 28$.
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