AMC 8 · 2020 · #23

Grade 7 counting

Archetype Overlap / Complement Counting └ Sub-archetype Two-Set Inclusion-Exclusion: Solve for Missing Region

combinations-basicpermutations-basiccomplementary-countingset-partition complementary-countingcasework ↑ Prerequisites: combinations-basicpermutations-basic

📏 Long solution 💡 4 insights

Problem

There are 5 different awards (think of them as 5 distinct trophies) and 3 different students. Every award must go to exactly one student, and every student must end up with at least one award. In how many different ways can this happen?

Pick an answer.

(A)

120

(B)

150

(C)

180

(D)

210

(E)

240

AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

The phrase "each student receives at least one award" is a textbook trigger for Tool #16 (Complement): instead of counting good distributions directly, we count ALL distributions and then subtract the bad ones (where at least one student gets nothing). Tool #7 (Identify Subproblems) helps because the "bad" count itself splits into clean pieces — first the distributions where a chosen student is empty, then a correction for distributions where two students are empty (these get double-subtracted and must be added back). Tool #3 (Eliminate) is the verification path: organize the 5 awards by the partition shape (3,1,1) or (2,2,1), count each case, and the two cases must add to the same total — eliminating the other choices.

1STEP 1

Ignore the rule first: each of 5 awards independently goes to any of 3 students, giving 3⁵ = 243 total distributions.

3⁵ = 243 total distributions

2STEP 2

Subtract the bad cases: pick the empty student (C(3,1)=3) and send all 5 awards to the other two (2⁵=32), giving 96 to remove.

C(3, 1) · 2⁵ = 3 · 32 = 96

3STEP 3

The two-empty cases were subtracted twice, so add them back: C(3,2)·1⁵ gives 3 cases to restore.

C(3, 2) · 1⁵ = 3 · 1 = 3

4STEP 4

Inclusion-exclusion: total − single-empty + double-empty = 243 − 96 + 3 = 150. No triple-empty term — all three empty is impossible.

3⁵ - C(3, 1) · 2⁵ + C(3, 2) · 1⁵ = 243 - 96 + 3 = 150 → (B)

Answer

150

Without any restriction the answer would be 3⁵ = 243. The restriction "each student gets at least one" rules out the lopsided distributions, so the answer must be smaller than 243. Among the choices only (A) 120, (B) 150, (C) 180, (D) 210 are below 243, and (E) 240 would mean almost no distributions were ruled out (unreasonable, since giving everything to one student is a big family of bad cases). Inclusion-Exclusion gives exactly 150, which sits comfortably below 243 and matches choice (B).

💡Key takeaway

This AMC 8 problem only needs Grade 7 compound-event counting — count all the ways, then take away the bad ones — that you already know!

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