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Difference between revisions of "2021 AMC 10A Problems/Problem 15"

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Assume that the first equation is above the second, since order doesn't matter. Then <math>C>A</math> and <math>B>D</math>. Therefore the number of ways to choose the four integers is <math>\tbinom{6}{2}\tbinom{4}{2}=90</math>.
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==Problem==
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Values for <math>A,B,C,</math> and <math>D</math> are to be selected from <math>\{1, 2, 3, 4, 5, 6\}</math> without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves <math>y=Ax^2+B</math> and <math>y=Cx^2+D</math> intersect? (The order in which the curves are listed does not matter; for example, the choices <math>A=3, B=2, C=4, D=1</math> is considered the same as the choices <math>A=4, B=1, C=3, D=2.</math>)
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<math>\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360</math>
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==Solution==
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Assume that the first equation is above the second, since order doesn't matter. Then <math>C>A</math> and <math>B>D</math>. Therefore the number of ways to choose the four integers is <math>\tbinom{6}{2}\tbinom{4}{2}=90</math>, and the answer is <math>\boxed{C}</math>.
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==See also==
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{{AMC10 box|year=2021|ab=A|num-b=14|num-a=16}}
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{{MAA Notice}}

Revision as of 14:02, 11 February 2021

Problem

Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$)

$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$

Solution

Assume that the first equation is above the second, since order doesn't matter. Then $C>A$ and $B>D$. Therefore the number of ways to choose the four integers is $\tbinom{6}{2}\tbinom{4}{2}=90$, and the answer is $\boxed{C}$.

See also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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