Difference between revisions of "2018 AMC 10B Problems/Problem 3"
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+ | == Problem == | ||
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In the expression <math>\left(\underline{\qquad}\times\underline{\qquad}\right)+\left(\underline{\qquad}\times\underline{\qquad}\right)</math> each blank is to be filled in with one of the digits <math>1,2,3,</math> or <math>4,</math> with each digit being used once. How many different values can be obtained? | In the expression <math>\left(\underline{\qquad}\times\underline{\qquad}\right)+\left(\underline{\qquad}\times\underline{\qquad}\right)</math> each blank is to be filled in with one of the digits <math>1,2,3,</math> or <math>4,</math> with each digit being used once. How many different values can be obtained? | ||
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\textbf{(E) }24 \qquad | \textbf{(E) }24 \qquad | ||
</math> | </math> | ||
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+ | == Solution == | ||
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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) | ||
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+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2018|ab=B|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Revision as of 15:02, 16 February 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 (harry1234)
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 |
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