AMC 8 · 2020 · #23
Grade 7 countingProblem
Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
Pick an answer.
AMC 8 2020 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: 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?
Givens: $5$ distinct awards (each award is unique); $3$ distinct students (the students are distinguishable from each other); Each award goes to exactly one student; Every student must receive at least $1$ award; Answer choices: (A) $120$, (B) $150$, (C) $180$, (D) $210$, (E) $240$
Unknowns: The number of distinct distributions of the $5$ awards to the $3$ students
Understand
Restated: 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?
Givens: $5$ distinct awards (each award is unique); $3$ distinct students (the students are distinguishable from each other); Each award goes to exactly one student; Every student must receive at least $1$ award; Answer choices: (A) $120$, (B) $150$, (C) $180$, (D) $210$, (E) $240$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
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.
Execute — Answer: B
6.EE.A.1 Step 1 - Count all distributions with NO restriction first.
- Each of the $5$ awards can go to any of the $3$ students independently, so by the multiplication principle the total is $3 \times 3 \times 3 \times 3 \times 3 = 3^5$.
💡 Writing $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3$ as $3^5$ is the Grade 6 whole-number exponent skill.
7.SP.C.8 Step 2 - Now subtract the BAD cases: distributions where at least one student gets zero awards.
- Pick which student is left out — there are $\binom{3}{1} = 3$ choices — and then send all $5$ awards to the other two students, which gives $2^5 = 32$ ways per choice.
💡 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.8 Step 3 - Correct the over-subtraction.
- The case where, say, both students $A$ and $B$ are empty was subtracted twice (once when we excluded $A$, once when we excluded $B$), so we must add it back.
- Choose which $2$ students are empty in $\binom{3}{2} = 3$ ways, and then all $5$ awards must go to the one remaining student — that is $1^5 = 1$ way.
💡 Recognizing and undoing double-counting is the Grade 7 reasoning step for compound events organized as overlapping cases.
7.SP.C.8 Step 4 - Combine the three numbers using the Inclusion-Exclusion pattern: total minus single-empty plus double-empty.
- (No need to subtract a triple-empty term because $3$ students all being empty is impossible — the awards have to go somewhere.)
💡 Adding and subtracting case counts in the right order is the Grade 7 method for counting compound events without double-counting.
6.EE.A.1 Count all distributions with NO restriction first. Each of the $5$ awards can go 7.SP.C.8 Now subtract the BAD cases: distributions where at least one student gets zero a 7.SP.C.8 Correct the over-subtraction. The case where, say, both students $A$ and $B$ are 7.SP.C.8 Combine the three numbers using the Inclusion-Exclusion pattern: total minus sin Review
Reasonableness: Without any restriction the answer would be $3^5 = 243$. 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).
Alternative: Tool #3 (Eliminate) via casework on the award-count pattern. Since each student needs $\ge 1$ award and the awards sum to $5$, the only two patterns are $(3,1,1)$ and $(2,2,1)$. Pattern $(3,1,1)$: pick the student who gets $3$ ($3$ ways), pick which $3$ of $5$ awards they get ($\binom{5}{3} = 10$ ways), then distribute the remaining $2$ awards to the other $2$ students ($2! = 2$ ways) — total $3 \cdot 10 \cdot 2 = 60$. Pattern $(2,2,1)$: pick the student who gets just $1$ ($3$ ways), pick which award they get ($5$ ways), then split the remaining $4$ awards into the two pairs for the other two students ($\binom{4}{2} = 6$ ways) — total $3 \cdot 5 \cdot 6 = 90$. Sum: $60 + 90 = 150$, again (B).
CCSS standards used (min grade 7)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Recognizing and computing $3^5 = 243$, $2^5 = 32$, $1^5 = 1$ from the multiplication principle.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting compound events (which student is excluded, which awards go where) and combining the case counts via the Inclusion-Exclusion pattern $243 - 96 + 3 = 150$.)
⭐ 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!
⭐ 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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