🇺🇸 AMC 8 ⇄ switch contest
2020 AMC 8

Problem 23

Problem 23 · 2020 AMC 8 Hard
Counting & Probability complementary-countingcasework

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?

Show answer
Answer: B — 150 ways.
Show hints
Hint 1 of 2
‘Each student gets at least one’ is the awkward part. Counting without that rule is easy — each of the 5 distinct awards independently goes to one of 3 students. So count freely, then remove the bad cases where someone is shut out.
Still stuck? Show hint 2 →
Hint 2 of 2
Total = 35. To remove the bad cases, use inclusion–exclusion: subtract the ‘a chosen student gets nothing’ cases (3 × 25), but you double-removed the ‘two students get nothing’ cases, so add those back (3 × 15).
Show solution
Approach: inclusion–exclusion on ‘someone is empty-handed’
  1. Awards are distinct and students are distinct, so without the ‘at least one’ rule each award has 3 independent choices: 35 = 243 total.
  2. Subtract the assignments where one named student gets nothing (the other two share all 5): C(3,1) × 25 = 3 × 32 = 96.
  3. Those subtractions double-counted the assignments where two named students get nothing (one student hogs everything), so add them back: C(3,2) × 15 = 3.
  4. 243 − 96 + 3 = 150.
  5. Why this transfers: ‘onto’ / ‘everyone gets something’ distributions are the home turf of inclusion–exclusion: count everything, subtract single exclusions, add back double exclusions. The alternating −, + pattern fixes the over-removal automatically.
Another way — split by the shape of the distribution:
  1. With 5 awards into 3 nonempty piles, the pile sizes are either 3-1-1 or 2-2-1 — the only ways to break 5 into three positive parts.
  2. Shape 3-1-1: pick who gets 3 (3 ways), choose their 3 awards C(5,3)=10, then the last 2 awards go one each to the other two students (2 ways): 3 × 10 × 2 = 60.
  3. Shape 2-2-1: pick who gets the single award (3 ways) and which award C(5,1)=5, then split the remaining 4 into two pairs for the two named students C(4,2)=6: 3 × 5 × 6 = 90.
  4. 60 + 90 = 150 — same answer, built directly with no subtracting.
Mark: · log in to save
↑