AMC 10 · 2005 · #18

Grade 7 arithmetic
Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements
combinations-basic systematic-enumerationcomplementary-counting ↑ Prerequisites: combinations-basic
📏 Short solution 💡 2 insights
Problem
A telephone number looks like 555-abc-defg, where the seven digits a,b,c,d,e,f,g are all different, are written in increasing order from left to right, and none of them is 0 or 1. Count how many such telephone numbers are possible.

Pick an answer.

(A)
1
(B)
2
(C)
7
(D)
8
(E)
9

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

The 'increasing order' rule is the hinge: it means arranging the digits is never a choice, so the only decision is which seven digits to use. That turns a scary-looking phone-number count into a plain 'choose a set of digits' count. Tool #3 (Eliminate Possibilities) first trims the digit pool by throwing out 0 and 1. Tool #16 (Count the Complement) then does the real work: instead of choosing the seven digits to keep, choose the one digit to leave out — a much smaller decision. Tool #2 (Make a Systematic List) is the fallback that confirms the same total by listing the choices directly.

1STEP 1

List the usable digits

Banning 0 and 1 leaves the pool 2,3,4,5,6,7,8,9 — eight usable digits for the seven slots.

{2,3,4,5,6,7,8,9} → 8 usable digits
2STEP 2

Increasing order fixes the arrangement

Increasing order allows exactly one arrangement per chosen set, so counting numbers is just counting seven-digit sets.

one chosen set of 7 digits ⟷ exactly one phone number
3STEP 3

Count by leaving one digit out

Keeping 7 of the 8 digits is the same as dropping exactly one, and there are 8 digits to drop — choice (D).

ways to keep 7 of 8 = ways to drop 1 of 8 = 8 → (D)
Answer
8
The answer 8 sits right inside the answer list and is small, which matches a problem where the arrangement is forced and only a handful of digit-sets exist. A quick sanity list confirms it: leaving out 2 gives 3456789, leaving out 3 gives 2456789, and so on through leaving out 9 to give 2345678 — exactly 8 numbers, one for each digit dropped. Choice (C) 7 is the trap for anyone who counts 'seven digits' instead of 'eight ways to choose them,' and (E) 9 traps anyone who forgets that 1 is banned and still counts nine digits 1 through 9.
💡Key takeaway

When the order is forced, you only count which digits to pick — and keeping 7 of 8 digits is the same as choosing the 1 to drop, so there are 8 numbers.

  • List the usable digits
  • Increasing order fixes the arrangement
  • Count by leaving one digit out

More like this

Same archetype — closest grade level first.