Competition · AMC preparation · step 4 of 4
AMC 8 · 2020 · #23
Grade 7 countingPick an answer.
AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Count all distributions
Ignore the rule first: each of 5 awards independently goes to any of 3 students, giving 3⁵ = 243 total distributions.
Writing 3 · 3 · 3 · 3 · 3 as 3⁵ is the Grade 6 whole-number exponent skill.
6.EE.A.1Change Focus Count The ComplementSubtract the empty-student cases
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.
Breaking "at least one empty" into "pick the empty student, then distribute" is exactly the Grade 7 organized-list approach to compound events.
7.SP.C.8Identify SubproblemsAdd back the double count
The two-empty cases were subtracted twice, so add them back: C(3,2)·1⁵ gives 3 cases to restore.
Recognizing and undoing double-counting is the Grade 7 reasoning step for compound events organized as overlapping cases.
7.SP.C.8Identify SubproblemsCombine by inclusion-exclusion
Inclusion-exclusion: total − single-empty + double-empty = 243 − 96 + 3 = 150. No triple-empty term — all three empty is impossible.
Adding and subtracting case counts in the right order is the Grade 7 method for counting compound events without double-counting.
The number of ways to hand out the awards so that every student gets at least one is 3⁵ - C(3, 1) · 2⁵ + C(3, 2) · 1⁵.
▸ Why?
The distributions where everyone gets an award are exactly all the distributions with the bad ones — those that leave some student empty — taken away.
▸ Why?
Every distribution either gives all three students something or leaves at least one student empty, and none is both, so the good count is the whole collection minus the bad collection.
▸ Why?
That whole collection is 3⁵: each of the 5 awards is handed to one of 3 students on its own, and those independent choices multiply, giving 3×3×3×3×3=3⁵.
▸ Why?
A distribution that leaves two students empty was removed once for each of those two students, so it was taken out twice; adding the two-empty count back one time cancels one of those removals so each bad distribution is removed exactly once.
▸ Why?
Subtracting the bad cases is only correct when each bad distribution is matched to a single removal, so a case that got removed twice must be paired back down to being removed once.
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!
- Count all distributions
- Subtract the empty-student cases
- Add back the double count
- Combine by inclusion-exclusion
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