Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Undo revision 137985 by Redfiretruck (talk))
(Tag: Undo)
(→‎Video Solution by Math-X (First understand the problem!!!))
 
(67 intermediate revisions by 27 users not shown)
Line 5: Line 5:
  
 
==Solution 1==
 
==Solution 1==
−
Write <math>S</math> and <math>T</math> for the Steelie and the Tiger respectively. Putting in <math>S</math> and <math>T</math> first, in order to avoid them being next to each other, we must have the arrangement <math>S\square T\square</math>, <math>S\square\square T</math>, <math>\square S\square T</math>, or any of these with the <math>S</math> and <math>T</math> swapped. This gives <math>3 \cdot 2 = 6</math> ways, and there are then <math>2</math> ways to put in the Aggie and Bumbleebee, for a total of <math>2\cdot 6=\boxed{\textbf{(C) }12}</math>.
+
Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by <math>A,B,S,</math> and <math>T</math>, respectively. If we 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\square\square, \square ST\square, \square\square ST</math>). However, we could also have placed <math>S</math> and <math>T</math> in the opposite order (i.e. <math>TS\square\square, \square TS\square, \square\square TS</math>). Thus there are 6 ways of placing <math>S</math> and <math>T</math> directly next to each other. Next, notice that for each of these placements, we have two open slots for placing <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}</math>.<br>
 +
 
 +
We can also solve this problem directly by looking at the number of ways that we can place <math>S</math> and <math>T</math> such that they are not directly next to each other. Observe that there are three ways to place <math>S</math> and <math>T</math> (in that order) into the four slots so they are not next to each other (i.e. <math>S\square T\square, \square S\square T, S\square\square T</math>). However, we could also have placed <math>S</math> and <math>T</math> in the opposite order (i.e. <math>T\square S\square, \square T\square S, T\square\square S</math>). Thus there are 6 ways of placing <math>S</math> and <math>T</math> so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing <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> such that they are not next to each other <math>\implies\boxed{\textbf{(C) }12}</math>.<br>
 +
~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
  
 
==Solution 2==
 
==Solution 2==
−
There would be <math>4!=24</math> ways to arrange the <math>4</math> marbles, except for the condition that the Steelie and Tiger cannot be next to each other. If we did place them next to each other with the Steelie first, there would be <math>3</math> ways to place them (namely <math>ST\square\square</math>, <math>\square ST\square</math>; or <math>\square\square ST</math>, where <math>S</math> and <math>T</math> denote the Steelie and the Tiger as in Solution 1). Accounting for the other possible order, there are a total of <math>3 \cdot 2 = 6</math> ways. Now, there are <math>2</math> ways to place <math>A</math> and <math>B</math>, giving overall <math>6 \cdot 2 = 12</math> ways to arrange the marbles with <math>S</math> and <math>T</math> next to each other. Subtracting this from <math>24</math> (to remove the cases which are not allowed) gives <math>24-12=\boxed{\textbf{(C) }12}</math> valid ways to arrange the marbles.
+
Let's try complementary counting. There <math>4!</math> ways to arrange the 4 marbles. However, there are <math>2\cdot3!</math> arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, <cmath>4!-2\cdot3!=\boxed{\textbf{(C) }12}</cmath>
 +
 
 +
==Solution 3==
 +
 
 +
We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a "super marble." There are <math>2!</math> ways to arrange Steelie and Tiger within this "super marble." Then there are <math>3!</math> ways to arrange the "super marble" and Zara's two other marbles in a row. Since there are <math>4!</math> ways to arrange the marbles without any restrictions, the answer is given by <math>4!-2!\cdot 3!=\boxed{\textbf{(C) }12}</math>
 +
 
 +
-franzliszt
 +
 
 +
==Solution 4(proof of Georgeooga-Harryooga Theorem used in solution 1)==
 +
 
 +
We will use the following
 +
 
 +
<math>\textbf{Georgeooga-Harryooga Theorem:}</math> The [[Georgeooga-Harryooga Theorem]] states that if you have <math>a</math> distinguishable objects and <math>b</math> of them cannot be together, then there are <math>\frac{(a-b)!(a-b+1)!}{b!}</math> ways to arrange the objects.
 +
 
 +
<math>\textit{Proof. (Created by AoPS user RedFireTruck)}</math>
 +
 
 +
Let our group of <math>a</math> objects be represented like so <math>1</math>, <math>2</math>, <math>3</math>, ..., <math>a-1</math>, <math>a</math>. Let the last <math>b</math> objects be the ones we can't have together.
 +
 
 +
Then we can organize our objects like so <math>\square1\square2\square3\square...\square a-b-1\square a-b\square</math>.
 +
 
 +
We have <math>(a-b)!</math> ways to arrange the objects in that list.
 +
 
 +
Now we have <math>a-b+1</math> blanks and <math>b</math> other objects so we have <math>_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}</math> ways to arrange the objects we can't put together.
 +
 
 +
By fundamental counting principle our answer is <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math>.
 +
 
 +
 
 +
Proof by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 12:09, 1 February 2021 (EST)
 +
 
 +
 
 +
Back to the problem. By the [[Georgeooga-Harryooga Theorem]], our answer is <math>\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{\textbf{(C) }12}</math>.
 +
 
 +
-franzliszt
 +
 
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
 +
 
 +
~NiuniuMaths
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=SRVyi268QzPhgWax&t=1232
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (🚀 Fast 🚀)==
 +
https://youtu.be/MC-KA1STkSY
 +
 
 +
~Education, the Study of Everything
  
−
==Solution 3 (variant of Solution 2)==
+
==Video Solution by STEMbreezy==
−
As in Solution 2, there are <math>24</math> total ways to arrange the marbles without any constraints. To count the number of ways where the Steelie and the Tiger are next to each other, we treat them together as a "super marble". There are <math>2</math> ways to arrange the Steelie and Tiger within the super marble, then <math>3! = 6</math> ways to put the super marble in a row with the Aggie and the Bumblebee. Thus the answer is <math>24-2\cdot 6=\boxed{\textbf{(C) }12}</math>.
+
https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=354
  
−
==Video Solution==
+
~STEMbreezy
−
https://youtu.be/pB46JzBNM6g
 
  
 
==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}}

Latest revision as of 06:18, 24 January 2024

Problem

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 1

Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$, respectively. If we 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. $ST\square\square, \square ST\square, \square\square ST$). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $TS\square\square, \square TS\square, \square\square TS$). 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 open slots for placing $A$ 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 $\implies\boxed{\textbf{(C) }12}$.

We can also solve this problem directly by looking at the number of ways that we can place $S$ and $T$ such that they are not directly next to each other. Observe that there are three ways to place $S$ and $T$ (in that order) into the four slots so they are not next to each other (i.e. $S\square T\square, \square S\square T, S\square\square T$). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $T\square S\square, \square T\square S, T\square\square S$). Thus there are 6 ways of placing $S$ and $T$ so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ 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 $T$ such that they are not next to each other $\implies\boxed{\textbf{(C) }12}$.
~junaidmansuri

Solution 2

Let's try complementary counting. There $4!$ ways to arrange the 4 marbles. However, there are $2\cdot3!$ arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, \[4!-2\cdot3!=\boxed{\textbf{(C) }12}\]

Solution 3

We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a "super marble." There are $2!$ ways to arrange Steelie and Tiger within this "super marble." Then there are $3!$ ways to arrange the "super marble" and Zara's two other marbles in a row. Since there are $4!$ ways to arrange the marbles without any restrictions, the answer is given by $4!-2!\cdot 3!=\boxed{\textbf{(C) }12}$

-franzliszt

Solution 4(proof of Georgeooga-Harryooga Theorem used in solution 1)

We will use the following

$\textbf{Georgeooga-Harryooga Theorem:}$ The Georgeooga-Harryooga Theorem states that if you have $a$ distinguishable objects and $b$ of them cannot be together, then there are $\frac{(a-b)!(a-b+1)!}{b!}$ ways to arrange the objects.

$\textit{Proof. (Created by AoPS user RedFireTruck)}$

Let our group of $a$ objects be represented like so $1$, $2$, $3$, ..., $a-1$, $a$. Let the last $b$ objects be the ones we can't have together.

Then we can organize our objects like so $\square1\square2\square3\square...\square a-b-1\square a-b\square$.

We have $(a-b)!$ ways to arrange the objects in that list.

Now we have $a-b+1$ blanks and $b$ other objects so we have $_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}$ ways to arrange the objects we can't put together.

By fundamental counting principle our answer is $\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$.


Proof by RedFireTruck (talk) 12:09, 1 February 2021 (EST)


Back to the problem. By the Georgeooga-Harryooga Theorem, our answer is $\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{\textbf{(C) }12}$.

-franzliszt

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=8hgK6rESdek&t=9s

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=SRVyi268QzPhgWax&t=1232

~Math-X

Video Solution (🚀 Fast 🚀)

https://youtu.be/MC-KA1STkSY

~Education, the Study of Everything

Video Solution by STEMbreezy

https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=354

~STEMbreezy

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_10&oldid=212549"