Difference between revisions of "2004 AIME II Problems/Problem 4"
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Now, let's count those with two distinct digits. We handle the cases "0 included" and "0 not included" separately. | 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. | + | 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,</math> <math>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. | 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. | ||
Revision as of 19:26, 4 March 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 | ||
