Difference between revisions of "2006 AMC 10A Problems/Problem 21"
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== Problem == | == Problem == | ||
− | How many four-digit positive integers have at least one digit that is a 2 or a 3? | + | How many four-digit positive integers have at least one digit that is a <math>2</math> or a <math>3</math>? |
− | <math>\ | + | <math>\textbf{(A) } 2439\qquad\textbf{(B) } 4096\qquad\textbf{(C) } 4903\qquad\textbf{(D) } 4904\qquad\textbf{(E) } 5416</math> |
− | == | + | |
− | == See | + | == Video Solution == |
− | + | https://youtu.be/0W3VmFp55cM?t=3291 | |
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Solution 1 (Complementary Counting) == | ||
+ | Since we are asked for the number of positive <math>4</math>-digit integers with at least <math>2</math> or <math>3</math> in it, we can find this by finding the total number of <math>4</math>-digit integers and subtracting off those which do not have any <math>2</math>s or <math>3</math>s as digits. | ||
+ | |||
+ | The total number of <math>4</math>-digit integers is <math>9 \cdot 10 \cdot 10 \cdot 10 = 9000</math>, since we have <math>10</math> choices for each digit except the first (which can't be <math>0</math>). | ||
+ | |||
+ | Similarly, the total number of <math>4</math>-digit integers without any <math>2</math> or <math>3</math> is <math>7 \cdot 8 \cdot 8 \cdot 8 ={3584}</math>. | ||
+ | |||
+ | Therefore, the total number of positive <math>4</math>-digit integers that have at least one <math>2</math> or <math>3</math> is <math>9000-3584=\boxed{\textbf{(E) }5416}.</math> | ||
+ | |||
+ | == Solution 2 (Casework)== | ||
+ | We proceed to every case. | ||
+ | |||
+ | Case <math>1</math>: There is ONLY one <math>2</math> or <math>3</math>. If the <math>2</math> or <math>3</math> is occupying the first digit, we have <math>512</math> arrangements. If the <math>2</math> or <math>3</math> is not occupying the first digit, there are <math>7 \cdot 8^2</math> = <math>448</math> arrangements. Therefore, we have <math>2(448 \cdot 3 + 512) = 3712</math> arrangements. | ||
+ | |||
+ | Case <math>2</math> : There are two <math>2</math>s OR two <math>3</math>s. If the <math>2</math> or <math>3</math> is occupying the first digit, we have <math>64</math> arrangements. If the <math>2</math> or <math>3</math> is not occupying the first digit, there are <math>56</math> arrangements. There are <math>3</math> ways for the <math>2</math> or the <math>3</math> to be occupying the first digit and <math>3</math> ways for the first digit to be unoccupied. There are <math>2(3 \cdot (56+64))</math> = <math>720</math> arrangements. | ||
+ | |||
+ | Case <math>3</math> : There is ONLY one <math>2</math> and one <math>3</math>. If the <math>2</math> or the <math>3</math> is occupying the first digit, we have <math>6</math> types of arrangements of where the <math>2</math> or <math>3</math> is. We also have <math>64</math> different arrangements for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 64</math> = <math>384</math> arrangements. If the <math>2</math> or the <math>3</math> isn't occupying the first digit, we have <math>6</math> types of arrangements of where the <math>2</math> or <math>3</math> is. We also have <math>56</math> different arrangements for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 56</math> = <math>336</math> arrangements for this case. We have <math>336 + 384</math> = <math>720</math> total arrangements for this case. | ||
+ | |||
+ | Notice that we already counted <math>3712 + 720 + 720 = 5152</math> 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 <math>\boxed{\textbf{(E) }5416}</math>. | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | == See also == | ||
+ | {{AMC10 box|year=2006|ab=A|num-b=20|num-a=22}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 09:40, 19 December 2021
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 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 |
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