AMC 10 · 2003 · #19

Grade 7 counting
Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements
permutations-basicprobability-basicfundamental-counting-principle complementary-countingidentify-subproblems ↑ Prerequisites: permutations-basic
📏 Medium solution 💡 2 insights
Problem
S is the set of all arrangements of 1,2,3,4,5 in a row whose first term is not 1. One arrangement is drawn from S at random, every member equally likely. The probability that its second term is 2, in lowest terms, is a/b — find a+b.

Pick an answer.

(A)
5
(B)
6
(C)
11
(D)
16
(E)
19
How to solve
Strategy Make a Systematic List

A probability over equally likely arrangements is a counting job, so Tool #2 (Make a Systematic List) is primary: fill the row position by position and multiply the number of choices at each position. Tool #16 (Change Focus / Count the Complement) sizes the sample space quickly — instead of counting arrangements that avoid 1 in front, count all 5! of them and subtract the 4! that do start with 1. Tool #7 (Identify Subproblems) keeps the numerator and the denominator as two separate counting questions, which is what stops the classic error of dividing by 120 after conditioning on S.

1STEP 1

Size the sample space

All arrangements number 120; removing those starting with 1 leaves 96.

|S| = 5! - 4! = 120 - 24 = 96
2STEP 2

Count arrangements with 2 second

Fixing the second position leaves 3 choices in front and 6 orders behind, giving 18.

3 × 1 × 3! = 3 × 1 × 6 = 18
3STEP 3

Form the probability and reduce

Dividing and reducing gives 3/16.

P = 18/96 = (18 ÷ 6)/(96 ÷ 6) = 3/16
4STEP 4

Read off the sum

Adding the two parts gives 19, choice (E).

a = 3, b = 16 → a + b = 19 → (E)
Answer
19
The value 3/16 = 0.1875 sits just under 1/5 = 0.2, and that is exactly right: banning 1 from the front pushes 1 into the other positions more often, so 1 takes more than its even share of position 2 and each of 2,3,4,5 takes slightly less. The four probabilities can be checked to total 1 with the probability for 1: arrangements with 1 second number 4 × 3! = 24, giving 24/96 = 1/4, and 1/4 + 4 × 3/16 = 4/16 + 12/16 = 1. The choices are built around two traps: answering 1/4 (thinking four numbers can sit second) gives 1+4=5, choice (A), and answering 1/5 (assuming plain symmetry) gives 1+5=6, choice (B).
💡Key takeaway

When a rule bans one number from one spot, that number crowds into the other spots, so everyone else's share of those spots drops a little below the even split.

  • Size the sample space
  • Count arrangements with 2 second
  • Form the probability and reduce
  • Read off the sum

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