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

Revision as of 23:15, 1 November 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 $\frac{6}{2}=\boxed{\textbf{(B) }3}$

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

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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

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

@@ Line 13: / Line 13: @@
 == 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{\textbf{(B) }3}</math>
 == 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 <math>(2, 3, 4)</math>.
+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{\textbf{(B) }3}</math> possible outcomes <math>(2, 3, 4)</math>.
 ==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{\textbf{(B) }3}</math>

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