Problem 4 · 2004 AMC 8
Easy
Counting & Probability
combinations
Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?
Show answer
Answer: B — 4.
Show hints
Hint 1 of 2
Choosing who plays is hard to picture, but flip it: picking 3 starters out of 4 is the same as picking the one person who sits out. How many ways to choose the one bench-warmer?
Still stuck? Show hint 2 →
Hint 2 of 2
This is complementary counting: count the small leftover group instead of the big chosen group. Choosing 3-of-4 and choosing 1-of-4 are the same count — that's the symmetry C(n, k) = C(n, n−k).
Show solution
Approach: count the one left out (complement)
- Each starting trio leaves exactly one person on the bench, so 'how many trios?' = 'how many ways to pick the one who sits?' There are 4 people, so 4 ways.
- Why this transfers: whenever you're choosing almost all of a group, count the tiny excluded part instead — far less work. Choosing 18 of 20, for instance, is just choosing the 2 to drop.
Another way — direct combination:
- C(4, 3) = 4!/(3! · 1!) = 4.
- Same 4, the long way.
Mark:
· log in to save