Difference between revisions of "2015 AMC 8 Problems/Problem 10"

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<math>\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561</math>
 
<math>\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561</math>
  
==Solution==
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==Solution 1==
  There are <math>9</math> choices for the first number, since it cannot be <math>0</math>, there are only <math>9</math> choices left for the second number since it must differ from the first, <math>8</math> choices for the third number, since it must differ from the first two, and <math>7</math> choices for the fourth number, since it must differ from all three.  This means there are <math>9 \times 9 \times 8 \times 7=\boxed{\textbf{(B) }4536}</math> integers between <math>1000</math> and <math>9999</math> with four distinct digits.
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There are <math>9</math> choices for the first number, since it cannot be <math>0</math>, there are only <math>9</math> choices left for the second number since it must differ from the first, <math>8</math> choices for the third number, since it must differ from the first two, and <math>7</math> choices for the fourth number, since it must differ from all three.  This means there are <math>9 \times 9 \times 8 \times 7=\boxed{\textbf{(B) }4536}</math> integers between <math>1000</math> and <math>9999</math> with four distinct digits.
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If you're too lazy to do <math>9 \times 9 \times 8 \times 7= 4536</math>, we can notice <math> 9 \times 9 = 81</math> and <math> 8 \times 7=56</math> The unit digit must be 6. Therefore the answer is the only answer with a a unit digit of 6. <math>\boxed{B}</math>)
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==Video Solution (HOW TO THINK CRITICALLY!!!)==
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https://youtu.be/-VcuRyDB_wI
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~Education, the Study of Everything
  
 
==Video solution==
 
==Video solution==
 
https://youtu.be/Zhsb5lv6jCI?t=272
 
https://youtu.be/Zhsb5lv6jCI?t=272
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https://www.youtube.com/watch?v=OESYIYjZFdk  ~David
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https://youtu.be/2nfFg8JXKFE
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~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2015|num-b=9|num-a=11}}
 
{{AMC8 box|year=2015|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:10, 18 August 2024

Problem

How many integers between $1000$ and $9999$ have four distinct digits?

$\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561$

Solution 1

There are $9$ choices for the first number, since it cannot be $0$, there are only $9$ choices left for the second number since it must differ from the first, $8$ choices for the third number, since it must differ from the first two, and $7$ choices for the fourth number, since it must differ from all three. This means there are $9 \times 9 \times 8 \times 7=\boxed{\textbf{(B) }4536}$ integers between $1000$ and $9999$ with four distinct digits.


If you're too lazy to do $9 \times 9 \times 8 \times 7= 4536$, we can notice $9 \times 9 = 81$ and $8 \times 7=56$ The unit digit must be 6. Therefore the answer is the only answer with a a unit digit of 6. $\boxed{B}$)

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/-VcuRyDB_wI

~Education, the Study of Everything

Video solution

https://youtu.be/Zhsb5lv6jCI?t=272

https://www.youtube.com/watch?v=OESYIYjZFdk ~David

https://youtu.be/2nfFg8JXKFE

~savannahsolver

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 AJHSME/AMC 8 Problems and Solutions

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