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Difference between revisions of "2024 AMC 10A Problems/Problem 6"

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== Problem ==
 
== Problem ==
 
What is the minimum number of successive swaps of adjacent letters in the string <math>ABCDEF</math> that are needed to change the string to <math>FEDCBA?</math> (For example, <math>3</math> swaps are required to change <math>ABC</math> to <math>CBA;</math> one such sequence of swaps is
 
What is the minimum number of successive swaps of adjacent letters in the string <math>ABCDEF</math> that are needed to change the string to <math>FEDCBA?</math> (For example, <math>3</math> swaps are required to change <math>ABC</math> to <math>CBA;</math> one such sequence of swaps is
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<math>ABC\rightarrow BAC\rightarrow BCA\rightarrow CBA.</math>)
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<math>ABC\to BAC\to BCA\to CBA.</math>)
  
 
<math>\textbf{(A)}~6\qquad\textbf{(B)}~10\qquad\textbf{(C)}~12\qquad\textbf{(D)}~15\qquad\textbf{(E)}~24</math>
 
<math>\textbf{(A)}~6\qquad\textbf{(B)}~10\qquad\textbf{(C)}~12\qquad\textbf{(D)}~15\qquad\textbf{(E)}~24</math>
  
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== Solution ==
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== Solution 1 (Analysis) ==
 
Procedurally, it takes:
 
Procedurally, it takes:
  
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There are <math>6</math> moves needed to change <math>ABCD</math> to <math>DCBA</math>.
 
There are <math>6</math> moves needed to change <math>ABCD</math> to <math>DCBA</math>.
  
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We see a pattern of <math>0,1,3,6,...</math>. We notice that the difference between consecutive terms is increasing by <math>1</math>, so in the same way, for <math>5</math> letters, we would need <math>10</math> moves, and for <math>6</math>, we would need <math>\boxed{\textbf{(D)} 15}</math> moves.
+
We see a pattern of <math>0,1,3,6,...</math>. We notice that the difference between consecutive terms is increasing by <math>1</math>, so in the same way, for <math>5</math> letters, we would need <math>10</math> moves, and for <math>6</math>, we would need <math>\boxed{\textbf{(D)}~15}</math> moves.
  
 
Thinking why, when we start making these moves, we see that for a string of length <math>n</math>, it takes <math>n-1</math> moves to move the last letter to the front. After, we get a string that will be changed identically to a string of length <math>n-1</math>. This works in our pattern above and is another way to think about the problem!
 
Thinking why, when we start making these moves, we see that for a string of length <math>n</math>, it takes <math>n-1</math> moves to move the last letter to the front. After, we get a string that will be changed identically to a string of length <math>n-1</math>. This works in our pattern above and is another way to think about the problem!
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~world123
 
~world123
  
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== Solution 3 (Solution 2 done fast)==
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Note:
  
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We can see that the most efficient way to change <math>ABCDEF</math> to <math>FEDCBA</math> is the same as changing <math>ABCDE</math> to <math>EBCDA</math> and then moving <math>F</math> to the front in <math>5</math> moves. Similarly, this would carry on downwards, where to change <math>ABCDE</math> to <math>EBCDA</math> would be to change <math>ABCD</math> to <math>DCBA</math> and move <math>E</math> <math>4</math> times to the front. This pattern will carry on until <math>AB</math> to <math>BA</math> would be <math>1</math>, and <math>A</math> to <math>A</math> would be <math>0</math>. The answer would be <math>0(A)+1(B)+2(C)+3(D)+4(E)+5(F)</math>, which is <math>\boxed{\textbf{(D)} 15}</math> moves.
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The sequence consists of triangular numbers shifted a term up (as it starts with <math>0</math> on term <math>1</math> and <math>1</math> on term <math>2</math>.)
 +
Thus, our explicit formula is <cmath>\dfrac{(n-1)(n+1-1)}{2} = \dfrac{(n)(n-1)}{2}</cmath> and as <math>n = 6</math> in this case (<math>6</math> letters), our answer is <math>\dfrac{(6)(6-1)}{2} = \boxed{\textbf{(D)}~15}</math>
 +
 
 +
~NSAoPS
 +
 
 +
== Solution 3 (Solution 2 Done Fast)==
 +
 
 +
We can see that the most efficient way to change <math>ABCDEF</math> to <math>FEDCBA</math> is the same as changing <math>ABCDE</math> to <math>EDCBA</math> and then moving <math>F</math> to the front in <math>5</math> moves. Similarly, this would carry on downwards, where to change <math>ABCDE</math> to <math>EDCBA</math> would be to change <math>ABCD</math> to <math>DCBA</math> and move <math>E</math> <math>4</math> times to the front. This pattern will carry on until <math>AB</math> to <math>BA</math> would be <math>1</math>, and <math>A</math> to <math>A</math> would be <math>0</math>. The answer would be <math>0(A)+1(B)+2(C)+3(D)+4(E)+5(F)</math>, which is <math>\boxed{\textbf{(D)}~15}</math> moves.
  
 
~Moonwatcher22
 
~Moonwatcher22
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 +
== Solution 4 (If you don't notice anything)==
 +
Simply, just blindly doing it, the most straightforward way is to shift F all the way back. From ABCDEF:
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<math>ABCDFE ->ABCFDE ->ABFCDE ->AFBCDE ->FABCDE</math>
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For E:
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<math>FABCED -> FABECD -> FAEBCD -> FEABCD</math>
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For D:
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<math>FEABDC -> FEADBC -> FEDABC</math>
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For C:
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<math>FEDACB -> FEDCAB</math>
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For B:
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FEDCBA
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That's it, you don't need to do it with A. Still <math>\boxed{\textbf{(D)}~15}</math> moves.
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 +
~pepper2831
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 +
==Solution 5 (Inversions of Permutations)==
 +
 +
Let the letters <math>A, B, C, D, E, F</math> correspond to the numbers <math>1, 2, 3, 4, 5, 6</math> respectively. We want to find the number of swaps required to transform the permutation <math>1 2 3 4 5 6</math> into the permutation <math>6 5 4 3 2 1</math>.
 +
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Here, we let <math>1 2 3 4 5 6</math> be the natural order. Then the total number of inversions of the permutation <math>6 5 4 3 2 1</math> is <math>1+2+3+4+5=15</math>. Hence the answer is <math>\boxed{\textbf{(D)}~15}</math>
 +
 +
~tsun26
 +
 +
== Video Solution by Pi Academy ==
 +
https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
  
 
== Video Solution 1 by Power Solve ==
 
== Video Solution 1 by Power Solve ==
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~Thesmartgreekmathdude
 
~Thesmartgreekmathdude
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 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=_o5zagJVe1U
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 +
==Video Solution by Just Math⚡==
 +
https://youtu.be/KrkjGpk1uZs
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=5|num-a=7}}
 
{{AMC10 box|year=2024|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:10, 24 November 2024

Problem

What is the minimum number of successive swaps of adjacent letters in the string $ABCDEF$ that are needed to change the string to $FEDCBA?$ (For example, $3$ swaps are required to change $ABC$ to $CBA;$ one such sequence of swaps is $ABC\to BAC\to BCA\to CBA.$)

$\textbf{(A)}~6\qquad\textbf{(B)}~10\qquad\textbf{(C)}~12\qquad\textbf{(D)}~15\qquad\textbf{(E)}~24$

Solution 1 (Analysis)

Procedurally, it takes:

  • $5$ swaps for $A$ to move to the sixth spot, giving $BCDEFA.$
  • $4$ swaps for $B$ to move to the fifth spot, giving $CDEFBA.$
  • $3$ swaps for $C$ to move to the fourth spot, giving $DEFCBA.$
  • $2$ swaps for $D$ to move to the third spot, giving $EFDCBA.$
  • $1$ swap for $E$ to move to the second spot (so $F$ becomes the first spot), giving $FEDCBA.$

Together, the answer is $5+4+3+2+1=\boxed{\textbf{(D)}~15}.$

~MRENTHUSIASM

Solution 2 (Recursive Approach)

We can proceed by a recursive tactic on the number of letters in the string.

Looking at the string $A$, there are $0$ moves needed to change it to the string $A$

Then, there is $1$ move to change $AB$ to $BA$.

Similarly, there is $3$ moves needed for three letters (said in the problem).

There are $6$ moves needed to change $ABCD$ to $DCBA$.

We see a pattern of $0,1,3,6,...$. We notice that the difference between consecutive terms is increasing by $1$, so in the same way, for $5$ letters, we would need $10$ moves, and for $6$, we would need $\boxed{\textbf{(D)}~15}$ moves.

Thinking why, when we start making these moves, we see that for a string of length $n$, it takes $n-1$ moves to move the last letter to the front. After, we get a string that will be changed identically to a string of length $n-1$. This works in our pattern above and is another way to think about the problem!

~world123

Note:

The sequence consists of triangular numbers shifted a term up (as it starts with $0$ on term $1$ and $1$ on term $2$.) Thus, our explicit formula is \[\dfrac{(n-1)(n+1-1)}{2} = \dfrac{(n)(n-1)}{2}\] and as $n = 6$ in this case ($6$ letters), our answer is $\dfrac{(6)(6-1)}{2} = \boxed{\textbf{(D)}~15}$

~NSAoPS

Solution 3 (Solution 2 Done Fast)

We can see that the most efficient way to change $ABCDEF$ to $FEDCBA$ is the same as changing $ABCDE$ to $EDCBA$ and then moving $F$ to the front in $5$ moves. Similarly, this would carry on downwards, where to change $ABCDE$ to $EDCBA$ would be to change $ABCD$ to $DCBA$ and move $E$ $4$ times to the front. This pattern will carry on until $AB$ to $BA$ would be $1$, and $A$ to $A$ would be $0$. The answer would be $0(A)+1(B)+2(C)+3(D)+4(E)+5(F)$, which is $\boxed{\textbf{(D)}~15}$ moves.

~Moonwatcher22

Solution 4 (If you don't notice anything)

Simply, just blindly doing it, the most straightforward way is to shift F all the way back. From ABCDEF:

$ABCDFE ->ABCFDE ->ABFCDE ->AFBCDE ->FABCDE$

For E:

$FABCED -> FABECD -> FAEBCD -> FEABCD$

For D:

$FEABDC -> FEADBC -> FEDABC$

For C:

$FEDACB -> FEDCAB$

For B:

FEDCBA

That's it, you don't need to do it with A. Still $\boxed{\textbf{(D)}~15}$ moves.

~pepper2831

Solution 5 (Inversions of Permutations)

Let the letters $A, B, C, D, E, F$ correspond to the numbers $1, 2, 3, 4, 5, 6$ respectively. We want to find the number of swaps required to transform the permutation $1 2 3 4 5 6$ into the permutation $6 5 4 3 2 1$.

Here, we let $1 2 3 4 5 6$ be the natural order. Then the total number of inversions of the permutation $6 5 4 3 2 1$ is $1+2+3+4+5=15$. Hence the answer is $\boxed{\textbf{(D)}~15}$

~tsun26

Video Solution by Pi Academy

https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv

Video Solution 1 by Power Solve

https://youtu.be/j-37jvqzhrg?si=ieBRx0-CUihcKttE&t=616

Video Solution by Daily Dose of Math

https://youtu.be/-EFTk2pBFug

~Thesmartgreekmathdude

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=_o5zagJVe1U

Video Solution by Just Math⚡

https://youtu.be/KrkjGpk1uZs

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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