2018 AMC 10B Problems/Problem 3

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 1 - Combinations

We have $\binom{4}{2}$ ways to choose the pairs, and we have $2!$ ways for the values to be rearranged, hence $\frac{6}{2}=\boxed{\textbf{(B) }3}$

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 $\boxed{\textbf{(B) }3}$ possible outcomes $(2, 3, 4)$ .

Solution 3

There are exactly $4!$ ways to arrange the numbers and $2!2!2!$ overcounts per way due to commutativity. Therefore, the answer is $\frac{4!}{2!2!2!}=\boxed{\textbf{(B) }3}$

~MC_ADe

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/clLyv4GFtto

~Education, the Study of Everything

Video Solution

https://youtu.be/KpC9wT7HgBo

~savannahsolver

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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