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

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== Solution ==
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== Solution 1 - Combinations ==
  
 
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>
 
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>
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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==
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https://youtu.be/KpC9wT7HgBo
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 13:07, 4 August 2020

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

https://youtu.be/KpC9wT7HgBo

~savannahsolver

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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