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

Latest revision as of 09:10, 18 August 2024

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 1

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.

If you're too lazy to do $9 \times 9 \times 8 \times 7= 4536$ , we can notice $9 \times 9 = 81$ and $8 \times 7=56$ The unit digit must be 6. Therefore the answer is the only answer with a a unit digit of 6. $\boxed{B}$ )

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/-VcuRyDB_wI

~Education, the Study of Everything

Video solution

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

https://www.youtube.com/watch?v=OESYIYjZFdk ~David

https://youtu.be/2nfFg8JXKFE

~savannahsolver

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.

@@ Line 1: / Line 1: @@
+==Problem==
 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>
 ==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.
+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.
+If you're too lazy to do <math>9 \times 9 \times 8 \times 7= 4536</math>, we can notice <math> 9 \times 9 = 81</math> and <math> 8 \times 7=56</math> The unit digit must be 6. Therefore the answer is the only answer with a a unit digit of 6. <math>\boxed{B}</math>)
+==Video Solution (HOW TO THINK CRITICALLY!!!)==
+https://youtu.be/-VcuRyDB_wI
+~Education, the Study of Everything
+==Video solution==
+https://youtu.be/Zhsb5lv6jCI?t=272
+https://www.youtube.com/watch?v=OESYIYjZFdk   ~David
+https://youtu.be/2nfFg8JXKFE
+~savannahsolver
+==See Also==
+{{AMC8 box|year=2015|num-b=9|num-a=11}}
+{{MAA Notice}}

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