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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 1==
 
==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.
+
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> integers between <math>1000</math> and <math>9999</math> with four distinct digits.
  
 
==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}}

Revision as of 13:04, 30 November 2015

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

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 $0$, $9$ choices 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.

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
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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