Difference between revisions of "2015 AMC 8 Problems/Problem 10"
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==Solution 1== | ==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. | 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. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2015|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:00, 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.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
