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

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==Problem==
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How many integers between <math>1000</math> and <math>9999</math> have four distinct digits?
 
How many integers between <math>1000</math> and <math>9999</math> have four distinct digits?
  
 
<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 1==
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The question can be rephrased to "How many four-digit positive integers have four distinct digits," since numbers between <math>1000</math> and <math>9999</math> are four-digit integers.  There are <math>9</math> choices for the first number, since it cannot be <math>0</math>, <math>9</math> choices 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> numbers between <math>1000</math> and <math>9999</math> with four distinct digits.
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==Solution==
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The question can be rephrased to "How many four-digit positive integers have four distinct digits?", since numbers between <math>1000</math> and <math>9999</math> are four-digit integers.  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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==Video solution==
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https://youtu.be/Zhsb5lv6jCI?t=272
  
 
==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 18:06, 16 January 2021

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

The question can be rephrased to "How many four-digit positive integers have four distinct digits?", since numbers between $1000$ and $9999$ are four-digit integers. 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.

Video solution

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

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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