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 | + | |
− | 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>, | + | ==Solution== |
+ | 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. | ||
+ | |||
+ | ==Video solution== | ||
+ | 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
Contents
Problem
How many integers between and have four distinct digits?
Solution
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 , there are only choices left 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 integers between and with four distinct digits.
Video solution
https://youtu.be/Zhsb5lv6jCI?t=272
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.