AMC 10 · 2010 · #22
Grade 7 countingSeven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?
Pick an answer.
AMC 10 2010 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: Seven different pieces of candy are placed into three bags — red, blue, and white. The red bag and the blue bag must each end up with at least one piece; the white bag is allowed to stay empty. Count how many different placements are possible.
Givens: There are $7$ distinct pieces of candy; There are $3$ bags: red, blue, white; The red bag must receive at least one piece; The blue bag must receive at least one piece; The white bag may be empty; Answer choices: (A) $1930$, (B) $1931$, (C) $1932$, (D) $1933$, (E) $1934$
Unknowns: The number of valid ways to place all $7$ candies into the three bags
Understand
Restated: Seven different pieces of candy are placed into three bags — red, blue, and white. The red bag and the blue bag must each end up with at least one piece; the white bag is allowed to stay empty. Count how many different placements are possible.
Givens: There are $7$ distinct pieces of candy; There are $3$ bags: red, blue, white; The red bag must receive at least one piece; The blue bag must receive at least one piece; The white bag may be empty; Answer choices: (A) $1930$, (B) $1931$, (C) $1932$, (D) $1933$, (E) $1934$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #12 Draw a Venn Diagram, #7 Identify Subproblems
The phrase "at least one" in two bags is the classic signal for Tool #16 (Count the Complement): counting placements that obey both rules head-on is messy, but counting ALL placements and then removing the rule-breaking ones is clean. The forbidden placements are "red empty" and "blue empty," and those two sets overlap, so Tool #12 (Venn / inclusion-exclusion) handles the double-count. Tool #7 (Identify Subproblems) splits the job into three simple power-of-a-number counts.
Execute — Answer: C
7.SP.C.8 Step 1 Count every placement first
- Ignore the rules for a moment.
- Each of the $7$ distinct candies independently picks one of the $3$ bags, so the number of placements is $3$ multiplied by itself $7$ times.
💡 It is easier to count the whole world of placements and then throw away the bad ones than to build only the good ones.
6.EE.A.1 Step 2 Count the red-empty placements
- A placement is illegal if the red bag is empty.
- If red must stay empty, every candy is forced into blue or white — only $2$ choices per candy.
- Evaluate that power of $2$.
💡 Banning one bag simply deletes one option from each candy, turning $3$ choices into $2$.
7.SP.C.8 Step 3 Count the blue-empty placements
- By the same reasoning, if the blue bag must stay empty, every candy goes into red or white — again $2$ choices each.
- The situation is a mirror image of the red case, so the count matches.
💡 Red and blue play identical roles in the rules, so their forbidden counts must be equal.
4.NBT.B.4 Step 4 Repair the overlap
- The red-empty pile and the blue-empty pile share some placements: the ones where red AND blue are both empty, meaning all $7$ candies sit in white.
- There is exactly $1$ such placement, and it was counted once in each pile, so it was removed twice.
- The true number of illegal placements is $128 + 128$ minus that $1$ double-counted case.
💡 The all-white placement got knocked out twice, so add it back once to keep the tally honest.
4.NBT.B.4 Step 5 Subtract the illegal placements
Take the full count of placements and remove the illegal ones to leave exactly the placements that give red and blue at least one candy each.
💡 Everything minus the rule-breakers is precisely the rule-followers — the answer is $1932$, choice (C).
7.SP.C.8 Ignore the rules for a moment. Each of the $7$ distinct candies independently pi 6.EE.A.1 A placement is illegal if the red bag is empty. If red must stay empty, every ca 7.SP.C.8 By the same reasoning, if the blue bag must stay empty, every candy goes into re 4.NBT.B.4 The red-empty pile and the blue-empty pile share some placements: the ones where 4.NBT.B.4 Take the full count of placements and remove the illegal ones to leave exactly t Review
Reasonableness: The answer choices sit within one unit of each other, so a divisibility check confirms the winner. The valid count must equal the number of ways to give each of the $7$ candies a bag with red and blue nonempty; every valid arrangement can be paired with the choice of which two candies are the "first" ones seeding red and blue, and the total is a multiple of $3$ (place a distinguished candy into one of the three bags as an outer loop). Only $1932$ is divisible by $3$ among the five choices ($1+9+3+2=15$, a multiple of $3$), while $1930, 1931, 1933, 1934$ are not. This matches the direct count $2187 - 255 = 1932$, choice (C).
Alternative: Split by how the white bag behaves using Tool #7. First distribute the $7$ candies into just red and blue with both nonempty: that is $2^{7} - 2 = 126$ ways (remove the two placements that dump everything into a single bag). Now allow white: instead, choose the subset going to white first ($2^{7} = 128$ subsets) and require the remaining candies to split between red and blue with both nonempty. Summing over subset sizes reproduces $2187 - 2\cdot 128 + 1 = 1932$, the same result reached without the complement shortcut.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Applying the counting (multiplication) principle to size the sample space: $3$ independent choices per candy give $3^{7}$ total placements, and $2$ choices per candy give the $2^{7}$ red-empty and blue-empty subcounts.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating the whole-number powers $3^{7} = 2187$ and $2^{7} = 128$ that the counting principle produces.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Combining the forbidden counts by inclusion-exclusion ($128 + 128 - 1 = 255$) and subtracting from the total ($2187 - 255 = 1932$).)
⭐ When two things each need "at least one," count every arrangement, subtract the ones where a bag is empty, and add back the case you removed twice.
⭐ When two things each need "at least one," count every arrangement, subtract the ones where a bag is empty, and add back the case you removed twice.
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