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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==
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{{AMC10 box|year=2018|ab=B|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 15:02, 16 February 2018

Problem

In the expression $\left(\underline{\qquad}\times\underline{\qquad}\right)+\left(\underline{\qquad}\times\underline{\qquad}\right)$ each blank is to be filled in with one of the digits $1,2,3,$ or $4,$ with each digit being used once. How many different values can be obtained?

$\textbf{(A) }2 \qquad \textbf{(B) }3\qquad \textbf{(C) }4 \qquad \textbf{(D) }6 \qquad \textbf{(E) }24 \qquad$

Solution

We have $\binom{4}{2}$ ways to choose the pairs, and we have $2$ ways for the values to be switched so $\frac{6}{2}=\boxed{3.}$ (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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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