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

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==Solution 3==
 
==Solution 3==
 
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>
 
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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==Video Solution (HOW TO THINK CRITICALLY!!!)==
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https://youtu.be/clLyv4GFtto
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~Education, the Study of Everything
  
  

Revision as of 12:31, 28 May 2023

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 switched so $\frac{6}{2}=\boxed{3.}$

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

Solution 3

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


Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/clLyv4GFtto

~Education, the Study of Everything


Video Solution

https://youtu.be/KpC9wT7HgBo

~savannahsolver

See Also

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

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