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Difference between revisions of "2006 AMC 10A Problems/Problem 21"

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== Solution (Casework)==
 
== Solution (Casework)==
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We proceed to the cases.
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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) = </math>3712<math> arrangements.
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Case </math>2<math>: There are Two </math>2<math>s or two </math>3<math>s but not both. 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)) = 720<math> arrangements.
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Case </math>3<math>: There is one </math>3<math> and one </math>2<math> but no more. 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.
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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) }
  
 
== See also ==
 
== See also ==

Revision as of 19:16, 25 October 2021

Problem

How many four-digit positive integers have at least one digit that is a $2$ or a $3$?

$\textbf{(A) } 2439\qquad\textbf{(B) } 4096\qquad\textbf{(C) } 4903\qquad\textbf{(D) } 4904\qquad\textbf{(E) } 5416$

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 $9 \cdot 10 \cdot 10 \cdot 10 = 9000$, 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 $7 \cdot 8 \cdot 8 \cdot 8 ={3584}$.

Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is $9000-3584=\boxed{5416} \Longrightarrow \mathrm{(E)}$

Solution (Casework)

We proceed to the cases.

Case $1$: There is only one $2$ or $3$. If the $2$ or $3$ is occupying the first digit, we have $512$ arrangements. If the $2$ or $3$ is not occupying the first digit, there are $7 \cdot 8^2$ = $448$ arrangements. Therefore, we have $2(448 \cdot 3 + 512) =$3712$arrangements.

Case$ (Error compiling LaTeX. Unknown error_msg)2$: There are Two$2$s or two$3$s but not both. If the$2$or$3$is occupying the first digit, we have$64$arrangements. If the$2$or$3$is not occupying the first digit, there are$56$arrangements. There are$3$ways for the$2$or the$3$to be occupying the first digit and$3$ways for the first digit to be unoccupied. There are$2(3 \cdot (56+64)) = 720$arrangements.

Case$ (Error compiling LaTeX. Unknown error_msg)3$: There is one$3$and one$2$but no more. If the$2$or the$3$is occupying the first digit, we have$6$types of arrangements of where the$2$or$3$is. We also have$64$different arrangements for the non-$2$or$3$digits. We have$6 \cdot 64$=$384$arrangements. If the$2$or the$3$isn't occupying the first digit, we have$6$types of arrangements of where the$2$or$3$is. We also have$56$different arrangements for the non-$2$or$3$digits. We have$6 \cdot 56$=$336$arrangements for this case. We have$336 + 384$=$720$total arrangements for this case.

Notice that we already counted$ (Error compiling LaTeX. Unknown error_msg)3712 + 720 + 720 = 5152$cases and we still have a lot of cases left over to count. This is already larger than the second largest answer choice, and therefore, our answer is$\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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