AMC 10 · 2010 · #22

Grade 7 counting
Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements
principle-of-inclusion-exclusioncomplementary-countingexponents complementary-counting ↑ Prerequisites: principle-of-inclusion-exclusion
📏 Medium solution 💡 3 insights
Problem
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.

Pick an answer.

(A)
1930
(B)
1931
(C)
1932
(D)
1933
(E)
1934

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

How to solve
Strategy Change Focus / Count the Complement

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.

1STEP 1

Count every placement first

Ignore the rules for a moment: each of the 7 candies picks one of 3 bags, giving 2187 placements in all.

3⁷ = 2187
2STEP 2

Count the red-empty placements

Red empty breaks the rule: every candy must then pick blue or white, just 2 options each, for 128 bad placements.

2⁷ = 128
3STEP 3

Count the blue-empty placements

Blue empty is the mirror image — each candy picks red or white — so that pile also holds 128 placements.

2⁷ = 128
4STEP 4

Repair the overlap

The two piles overlap in one placement — all 7 candies in white — so the illegal total is 255, not 256.

128 + 128 - 1 = 255
5STEP 5

Subtract the illegal placements

Subtract the rule-breakers from the full count, and only the placements giving red and blue at least one candy remain.

2187 - 255 = 1932 → (C)
Answer
1932
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).
💡Key takeaway

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.

  • Count every placement first
  • Count the red-empty placements
  • Count the blue-empty placements
  • Repair the overlap
  • Subtract the illegal placements

More like this

Same archetype — closest grade level first.