Difference between revisions of "2004 AIME II Problems/Problem 4"
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== Problem == | == Problem == | ||
− | How many positive | + | How many [[positive integer]]s less than 10,000 have at most two different [[digit]]s? |
== Solution == | == 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 <math>9 \choose 2</math> ways to choose two digits, <math>A</math> and <math>B</math>. Given two digits, there are <math>2^n - 2</math> ways to arrange them in an <math>n</math>-digit number, for a total of <math>(2^1 - 2) + (2^2 - 2) + (2^3 -2) + (2^4 - 2) = 22</math> such numbers (or we can list them: <math>AB, BA, AAB, ABA, BAA, ABB, BAB, BBA, AAAB, AABA, ABAA, BAAA, AABB, ABAB, BAAB, ABBA, BABA, BBAA, ABBB, BABB, BBAB, BBBA</math>). Thus, we have <math>{9 \choose 2} \cdot 22 = 36\cdot22 = 792</math> 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 <math>2^{n - 1} - 1</math> <math>n</math>-digit numbers we can form, for a total of <math>(2^0 - 1) + (2^1 - 1) + (2^2 - 1) + (2^3 - 1) = 11</math> such numbers (or we can list them: <math>A0, A00, A0A, AA0, A000, AA00, A0A0, A00A, AAA0, AA0A, A0AA</math>). This gives us <math>9\cdot 11 = 99</math> numbers of this form. | ||
+ | |||
+ | Thus, in total, we have <math>36 + 792 + 99 = 927</math> such numbers. | ||
+ | |||
== See also == | == See also == | ||
* [[2004 AIME II Problems/Problem 3 | Previous problem]] | * [[2004 AIME II Problems/Problem 3 | Previous problem]] | ||
* [[2004 AIME II Problems/Problem 5 | Next problem]] | * [[2004 AIME II Problems/Problem 5 | Next problem]] | ||
* [[2004 AIME II Problems]] | * [[2004 AIME II Problems]] | ||
+ | [[Category:Intermediate Combinatorics Problems]] |
Revision as of 21:23, 12 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.