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_3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229
Solution 1 (Complementary Counting)
Since we are asked for the number of positive -digit integers with at least or in it, we can find this by finding the total number of -digit integers and subtracting off those which do not have any s or s as digits.
The total number of -digit integers is , since we have choices for each digit except the first (which can't be ).
Similarly, the total number of -digit integers without any or is .
Therefore, the total number of positive -digit integers that have at least one or is
Solution 2 (Casework)
We proceed to every case.
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 arrangements.
Case : There are two s OR two s. If the or is occupying the first digit, we have arrangements. If the or is not occupying the first digit, there are arrangements. There are ways for the or the to be occupying the first digit and ways for the first digit to be unoccupied. There are = arrangements.
Case : There is ONLY one and one . If the or the is occupying the first digit, we have types of arrangements of where the or is. We also have different arrangements for the non- or digits. We have = arrangements. If the or the isn't occupying the first digit, we have types of arrangements of where the or is. We also have different arrangements for the non- or digits. We have = arrangements for this case. We have = total arrangements for this case.
Notice that we already counted cases and we still have a lot of cases left over to count. This is already larger than the second largest answer choice, and therefore, our answer is .
~Arcticturn
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.