Difference between revisions of "2018 AMC 10B Problems/Problem 3"
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== 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. | + | 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>. |
==See Also== | ==See Also== |
Revision as of 15:07, 16 February 2018
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
We have ways to choose the pairs, and we have ways for the values to be switched so (harry1234)
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 .
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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All AMC 10 Problems and Solutions |
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