Difference between revisions of "2018 AMC 10B Problems/Problem 3"

(Solution)
(Solution 2)
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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.
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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>.
  
 
==See Also==
 
==See Also==

Revision as of 15:07, 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)

Solution 2

We have four available numbers $(1, 2, 3, 4)$. Because different permutations do not matter because they are all addition and multiplication, if we put $1$ on the first space, it is obvious there are $\boxed{3}$ possible outcomes $(2, 3, 4)$.

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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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