Difference between revisions of "2004 AIME II Problems/Problem 4"
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] |
Revision as of 21:51, 13 February 2007
Problem
How many positive integers less than 10,000 have at most two different digits?
Solution
First, let's count numbers with only a single digit. We have nine of these for each length, and four lengths, so 36 total numbers.
Now, let's count those with two distinct digits. We handle the cases "0 included" and "0 not included" separately.
There are ways to choose two digits,
and
. Given two digits, there are
ways to arrange them in an
-digit number, for a total of
such numbers (or we can list them:
). Thus, we have
numbers of this form.
Now, suppose 0 is one of our digits. We have nine choices for the other digit. For each choice, we have
-digit numbers we can form, for a total of
such numbers (or we can list them:
). This gives us
numbers of this form.
Thus, in total, we have such numbers.
See also
2004 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |