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Difference between revisions of "2004 AIME II Problems/Problem 4"

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(This isn't a very good solution. Anyone want to fix?)
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== Problem ==
 
== Problem ==
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How many positive integers less than 10,000 have at most two different digits?
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How many [[positive integer]]s less than 10,000 have at most two different [[digit]]s?
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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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.
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Now, let's count those with two distinct digits.  We handle the cases "0 included" and "0 not included" separately.
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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.
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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.
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Thus, in total, we have <math>36 + 792 + 99 = 927</math> such numbers.
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== 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]]
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 20:23, 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 $9 \choose 2$ ways to choose two digits, $A$ and $B$. Given two digits, there are $2^n - 2$ ways to arrange them in an $n$-digit number, for a total of $(2^1 - 2) + (2^2 - 2) + (2^3 -2) + (2^4 - 2) = 22$ such numbers (or we can list them: $AB, BA, AAB, ABA, BAA, ABB, BAB, BBA, AAAB, AABA, ABAA, BAAA, AABB, ABAB, BAAB, ABBA, BABA, BBAA, ABBB, BABB, BBAB, BBBA$). Thus, we have ${9 \choose 2} \cdot 22 = 36\cdot22 = 792$ 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 $2^{n - 1} - 1$ $n$-digit numbers we can form, for a total of $(2^0 - 1) + (2^1 - 1) + (2^2 - 1) + (2^3 - 1) = 11$ such numbers (or we can list them: $A0, A00, A0A, AA0, A000, AA00, A0A0, A00A, AAA0, AA0A, A0AA$). This gives us $9\cdot 11 = 99$ numbers of this form.

Thus, in total, we have $36 + 792 + 99 = 927$ such numbers.

See also

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