Difference between revisions of "2018 AMC 10B Problems/Problem 3"
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We have <math>\binom{4}{2}</math> ways to choose the pairs, and we have <math>2</math> ways for the values to be switched so <math>\frac{6}{2}=\boxed{3.}</math> (harry1234) | We have <math>\binom{4}{2}</math> ways to choose the pairs, and we have <math>2</math> ways for the values to be switched so <math>\frac{6}{2}=\boxed{3.}</math> (harry1234) | ||
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| + | == Solution 2 == | ||
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| + | 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. | ||
==See Also== | ==See Also== | ||
Revision as of 14:07, 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 | |
| 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. 
