2006 AMC 10A Problems/Problem 21
Contents
Problem
How many four-digit positive integers have at least one digit that is a or a
?
Video Solution
https://youtu.be/0W3VmFp55cM?t=3291
~ pi_is_3.14
Solution (Complementary Counting)
Since we are asked for the number of positive 4-digit integers with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits.
The total number of 4-digit integers is , since we have 10 choices for each digit except the first (which can't be 0).
Similarly, the total number of 4-digit integers without any 2 or 3 is .
Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is
Solution (Casework)
We proceed to the cases.
Case : There is only one
or
. If the
or
is occupying the first digit, we have
arrangements. If the
or
is not occupying the first digit, there are
=
arrangements. Therefore, we have
3712$arrangements.
Case$ (Error compiling LaTeX. Unknown error_msg)22
3
2
3
64
2
3
56
3
2
3
3
2(3 \cdot (56+64)) = 720$arrangements.
Case$ (Error compiling LaTeX. Unknown error_msg)33
2
2
3
6
2
3
64
2
3
6 \cdot 64
384
2
3
6
2
3
56
2
3
6 \cdot 56
336
336 + 384
720$total arrangements for this case.
Notice that we already counted$ (Error compiling LaTeX. Unknown error_msg)3712 + 720 + 720 = 5152\boxed{\textbf{(E) }
See also
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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