2015 AMC 8 Problems/Problem 10: Difference between revisions
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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== | |||
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. | |||
Revision as of 15:58, 25 November 2015
How many integers between 
and 
have four distinct digits?
Solution 1
The question can be rephrased to "How many four-digit positive integers have four distinct digits," since numbers between and are four-digit integers. There are 
choices for the first number, since it cannot be 
, choices for the second number, since it must differ from the first, 
choices for the third number, since it must differ from the first two, and 
choices for the fourth number, since it must differ from all three. This means there are 
numbers between and with four distinct digits.
