Difference between revisions of "2018 AMC 10B Problems/Problem 3"
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| − | == Solution == | + | == Solution 1 - Combinations == |
| − | We have <math>\binom{4}{2}</math> ways to choose the pairs, and we have <math>2</math> ways for the values to be | + | We have <math>\binom{4}{2}</math> ways to choose the pairs, and we have <math>2!</math> ways for the values to be rearranged, hence <math>\frac{6}{2}=\boxed{\textbf{(B) }3}</math> |
== Solution 2 == | == Solution 2 == | ||
| − | We have four available numbers <math>(1, 2, 3, 4)</math>. Because different permutations do not matter because they are all addition and multiplication, if we put <math>1</math> on the first space, it is obvious there are <math>\boxed{3}</math> possible outcomes <math>(2, 3, 4)</math>. | + | We have four available numbers <math>(1, 2, 3, 4)</math>. Because different permutations do not matter because they are all addition and multiplication, if we put <math>1</math> on the first space, it is obvious there are <math>\boxed{\textbf{(B) }3}</math> possible outcomes <math>(2, 3, 4)</math>. |
| − | ==Solution | + | ==Solution 3== |
| − | There are <math>4!</math> ways to arrange the numbers and <math>2!2!2!</math> overcounts per way due to commutativity. Therefore, the answer is <math>\frac{4!}{2!2!2!}=\boxed{3}</math> | + | There are exactly <math>4!</math> ways to arrange the numbers and <math>2!2!2!</math> overcounts per way due to commutativity. Therefore, the answer is <math>\frac{4!}{2!2!2!}=\boxed{\textbf{(B) }3}</math> |
| + | ~MC_ADe | ||
| + | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
| + | https://youtu.be/clLyv4GFtto | ||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/KpC9wT7HgBo | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Latest revision as of 21:27, 10 November 2023
Contents
Problem
In the expression 
each blank is to be filled in with one of the digits 
or 
with each digit being used once. How many different values can be obtained?
Solution 1 - Combinations
We have 
ways to choose the pairs, and we have 
ways for the values to be rearranged, hence 
Solution 2
We have four available numbers 
. Because different permutations do not matter because they are all addition and multiplication, if we put 
on the first space, it is obvious there are 
possible outcomes 
.
Solution 3
There are exactly 
ways to arrange the numbers and 
overcounts per way due to commutativity. Therefore, the answer is 
~MC_ADe
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2018 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
