Difference between revisions of "2018 AMC 10B Problems/Problem 3"
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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 <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{3}</math> possible outcomes <math>(2, 3, 4)</math>. | ||
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+ | ==Solution 4== | ||
+ | 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> | ||
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==See Also== | ==See Also== |
Revision as of 16:59, 19 March 2018
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
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 4
There are ways to arrange the numbers and
overcounts per way due to commutativity. Therefore, the answer is
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.