2005 AMC 10B Problem 18
All of David's telephone numbers have the form , where , , , , , , and are distinct digits and in increasing order, and none is either or . How many different telephone numbers can David have?
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Key insight
Seven different digits in increasing order can be arranged only one way. So a phone number is fixed by which of the eight digits 2 through 9 is left out.
Solution
- Which digits are allowed? Not or , so only . That is digits.
- How many are used? The letters through are different digits. So exactly one of the allowed digits is not used.
- Order is decided for us. Once you know which digits are used, "increasing order" leaves just one way to write them. For example, if is left out, the number must be --.
- Count. So each phone number matches exactly one choice of the digit to leave out. There are digits that could be left out, so there are phone numbers.
The answer is .
Why this works
"Different digits in increasing order" means the order is not a choice; only the set of digits matters. So count the sets, not the arrangements. Choosing of digits is the same as choosing the digit to leave out, which is much easier to count.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
The allowed digits are : eight of them. The seven positions through must be filled with seven distinct allowed digits.
Once we decide which seven digits to use, the increasing-order requirement leaves only one way to place them. So a telephone number corresponds exactly to a choice of digits out of , or equivalently to the one digit that is omitted.
There are digits that could be the omitted one, so there are telephone numbers.
The answer is .
Review in the AoPS books
- Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
- Intermediate Counting & Probability, Chapter 4: Constructive Counting and 1-1 Correspondences (§4.4, §4.5, §4.6)
- Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
- Introduction to Counting & Probability, Chapter 4: Committees and Combinations
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.