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Difference between revisions of "2020 AMC 8 Problems/Problem 10"

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<math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math>
 
<math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math>
 
==Solution==
 
==Solution==
−
Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by <math>A,B,S,</math> and <math>T</math>, respectively. If the ignore the constraint that <math>S</math> and <math>T</math> cannot be next to each other, we get a total of <math>4!=24</math> ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that <math>S</math> and <math>T</math> can be next to each other. If we place <math>S</math> and <math>T</math> next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. <math>ST\fbox{}\fbox{}</math>). However, we could also have placed <math>S</math> and <math>T</math> in the opposite order (i.e. <math>TS\fbox\fbox</math>). Thus there are 6 ways of placing <math>S</math> and <math>T directly next to each other. Next, notice that for each of these placements, we have two empty slots for placing </math>A<math> and </math>B<math>. Thus, we have 2 ways to place </math>A<math> and </math>B<math>. Specifically, we can place </math>A<math> in the first open slot and </math>B<math> in the second open slot or switch their order and place </math>B<math> in the first open slot and </math>A<math> in the second open slot. This gives us a total of </math>6\times 2=12<math> ways to place </math>S<math> and </math>T<math> next to each other. Subtracting this from the total number of arrangements gives us </math>24-12=12<math> total arrangements </math>\implies\boxed{\textbf{(C) }12}$.
+
Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by <math>A,B,S,</math> and <math>T</math>, respectively. If the ignore the constraint that <math>S</math> and <math>T</math> cannot be next to each other, we get a total of <math>4!=24</math> ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that <math>S</math> and <math>T</math> can be next to each other. If we place <math>S</math> and <math>T</math> next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. <math>ST\enclosesqaure</math>). However, we could also have placed <math>S</math> and <math>T</math> in the opposite order (i.e. <math>TS</math>). Thus there are 6 ways of placing <math>S</math> and <math>T directly next to each other. Next, notice that for each of these placements, we have two empty slots for placing </math>A<math> and </math>B<math>. Thus, we have 2 ways to place </math>A<math> and </math>B<math>. Specifically, we can place </math>A<math> in the first open slot and </math>B<math> in the second open slot or switch their order and place </math>B<math> in the first open slot and </math>A<math> in the second open slot. This gives us a total of </math>6\times 2=12<math> ways to place </math>S<math> and </math>T<math> next to each other. Subtracting this from the total number of arrangements gives us </math>24-12=12<math> total arrangements </math>\implies\boxed{\textbf{(C) }12}$.
  
 
==See also==
 
==See also==
 
{{AMC8 box|year=2020|num-b=9|num-a=11}}
 
{{AMC8 box|year=2020|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:21, 18 November 2020

Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?

$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution

Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$, respectively. If the ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each other. If we place $S$ and $T$ next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. $ST\enclosesqaure$ (Error compiling LaTeX. Unknown error_msg)). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $TS$). Thus there are 6 ways of placing $S$ and $T directly next to each other. Next, notice that for each of these placements, we have two empty slots for placing$A$and$B$. Thus, we have 2 ways to place$A$and$B$. Specifically, we can place$A$in the first open slot and$B$in the second open slot or switch their order and place$B$in the first open slot and$A$in the second open slot. This gives us a total of$6\times 2=12$ways to place$S$and$T$next to each other. Subtracting this from the total number of arrangements gives us$24-12=12$total arrangements$\implies\boxed{\textbf{(C) }12}$.

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 AJHSME/AMC 8 Problems and Solutions

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