Difference between revisions of "2012 AMC 8 Problems/Problem 10"
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<math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12 </math> | <math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of <math> 2012 </math>, since all of the valid 4-digit number will always be greater than <math> 1000 </math>. The best way to solve this problem is by using casework. | For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of <math> 2012 </math>, since all of the valid 4-digit number will always be greater than <math> 1000 </math>. The best way to solve this problem is by using casework. | ||
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Summing the amounts of numbers, we find that there are <math> \boxed{\textbf{(D)}\ 9} </math> such numbers. | Summing the amounts of numbers, we find that there are <math> \boxed{\textbf{(D)}\ 9} </math> such numbers. | ||
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| + | ==Solution 2== | ||
| + | Notice that the first digit cannot be <math> 0 </math>, as the number is greater than <math> 1000 </math>. Therefore, there are three digits that can be in the thousands. | ||
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| + | The rest three digits of the number have no restrictions, and therefore there are <math> 3! \implies 6 </math> for each leading digit. | ||
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| + | Since the two <math> 2 </math>'s are indistinguishable, there are <math> \frac {3\cdot6}{2} </math> such numbers <math> \implies \boxed{\textbf{(D)}\ 9} </math>. | ||
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| + | ==Solution 3 (a simpler version of Solution 2)== | ||
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| + | We can list out the four digits in the number. The first digit of the number can’t be 0 since the number is greater than <math>1000</math>. This leaves us with three integers for the digit in the thousands place. We have no restrictions in the hundreds place except for the fact that it can’t be the integer we chose for the thousands place. Continuing this pattern, there are <math>3</math> choices for the thousands place, <math>3</math> for the hundreds, <math>2</math> for the tens, and <math>1</math> for the ones. Adding this up we get <math>9</math> which is choice <math>\boxed{\textbf{(D)} }</math>. | ||
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| + | —-jason.ca | ||
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| + | ==Solution 4 (Complementary Counting)== | ||
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| + | There are <math>\frac{4!}{2!}</math> ways to arrange the numbers in <math>2012</math>. The number of ways to arrange 2012 with 0 as the first digit is 3 because there are three places to put the 1 and the 2s are the same. <math>12-3= 9</math> which is choice <math>\boxed{\textbf{(D)} }</math>. | ||
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| + | ~ Edited by GeometryMystery | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/OBKIeTgw4Zg ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=9|num-a=11}} | {{AMC8 box|year=2012|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:00, 6 July 2024
Contents
Problem
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
Solution 1
For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of 
, since all of the valid 4-digit number will always be greater than 
. The best way to solve this problem is by using casework.
There can be only two leading digits, namely 
or 
.
When the leading digit is , you can make 
such numbers.
When the leading digit is , you can make 
such numbers.
Summing the amounts of numbers, we find that there are 
such numbers.
Solution 2
Notice that the first digit cannot be 
, as the number is greater than . Therefore, there are three digits that can be in the thousands.
The rest three digits of the number have no restrictions, and therefore there are for each leading digit.
Since the two 's are indistinguishable, there are 
such numbers 
.
Solution 3 (a simpler version of Solution 2)
We can list out the four digits in the number. The first digit of the number can’t be 0 since the number is greater than . This leaves us with three integers for the digit in the thousands place. We have no restrictions in the hundreds place except for the fact that it can’t be the integer we chose for the thousands place. Continuing this pattern, there are 
choices for the thousands place, for the hundreds, for the tens, and for the ones. Adding this up we get 
which is choice 
.
—-jason.ca
Solution 4 (Complementary Counting)
There are 
ways to arrange the numbers in . The number of ways to arrange 2012 with 0 as the first digit is 3 because there are three places to put the 1 and the 2s are the same. 
which is choice .
~ Edited by GeometryMystery
Video Solution
https://youtu.be/OBKIeTgw4Zg ~savannahsolver
See Also
| 2012 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. 
