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Difference between revisions of "2004 AIME II Problems/Problem 4"
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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.
 
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.
+
Thus, in total, we have <math>36 + 792 + 99 = \boxed{927}</math> such numbers.
  
 
== See also ==
 
== See also ==

Revision as of 17:47, 20 July 2017

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 $36 + 792 + 99 = \boxed{927}$ 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

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