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

 
(Solution (Casework))
 
(27 intermediate revisions by 16 users not shown)
Line 1: Line 1:
 
== 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>\mathrm{(A) \ } 2439\qquad\mathrm{(B) \ } 4096\qquad\mathrm{(C) \ } 4903\qquad\mathrm{(D) \ } 4904\qquad\mathrm{(E) \ } 5416\qquad</math>
+
<math>\textbf{(A) } 2439\qquad\textbf{(B) } 4096\qquad\textbf{(C) } 4903\qquad\textbf{(D) } 4904\qquad\textbf{(E) } 5416</math>
== Solution ==
+
 
== See Also ==
+
== Video Solution ==
*[[2006 AMC 10A Problems]]
+
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

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 1 (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 $2$s or $3$s 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 $3$ is $9000-3584=\boxed{\textbf{(E) }5416}.$

Solution 2 (Casework)

We proceed to every case.

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 $2$ : There are two $2$s OR two $3$s. 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 $3$ : There is ONLY one $2$ and one $3$. 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 $3712 + 720 + 720 = 5152$ 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 $\boxed{\textbf{(E) }5416}$.

~Arcticturn

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS