AMC 10 · 2012 · #16

Grade 8 countinglogic
combinations-basiccomplementary-counting casework ↑ Prerequisites: combinations-basic
📏 Medium solution 💡 3 insights
Problem
No item may be liked by everyone, and every pair must share at least one item. Count the arrangements.

Pick an answer.

(A)
108
(B)
132
(C)
671
(D)
846
(E)
1105
How to solve
Strategy Change Focus / Count the Complement

The three pair conditions are all 'at least one' requirements, which are awkward to count head-on. First reframe each song by the set of girls who like it and list the allowed sets. Then, instead of building only the good arrangements, count them by complement: start from every arrangement and use inclusion-exclusion to subtract the ones that miss a required pair.

1STEP 1

List each song's possible type

Each item carries one of seven labels.

7 types: ∅, {A},{B},{J}, {A,B},{B,J},{A,J}
2STEP 2

Restate the pair rule as coverage

The pair rule says three labels must all appear.

3STEP 3

Count all label assignments first

Counting every assignment is easy.

7⁴ = 2401
4STEP 4

Subtract arrangements missing a pair

Subtracting the misses alternates in sign.

7⁴ - C(3, 1)6⁴ + C(3, 2)5⁴ - C(3, 3)4⁴
5STEP 5

Compute the final total

The total is 132, choice (B).

2401 - 3888 + 1875 - 256 = 132
Answer
132
132 is far smaller than the 2401 unrestricted arrangements, which fits: forcing all three pair-types to appear is a strong restriction. It also comfortably exceeds the 24 ways of merely seating the three pair-types on three songs, since the fourth song adds several options. Among the choices only (B) 132 lands in this range; the large options 671, 846, and 1105 would require almost no restriction at all.
💡Key takeaway

When a problem demands 'at least one of each,' count every arrangement first and then subtract the ones that leave something out.

  • List each song's possible type
  • Restate the pair rule as coverage
  • Count all label assignments first
  • Subtract arrangements missing a pair
  • Compute the final total