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Difference between revisions of "2010 AMC 8 Problems/Problem 25"

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==Solution 1==
 
==Solution 1==
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A valid climb is a sequence of some or all of the <math>1</math>, <math>2</math>, and <math>3</math> step hops, in which the sum of the sequence adds to <math>6</math>. There are three possible sequences which only contain one number- all the <math>1s</math>, all <math>2s</math>, or all <math>3s</math>. If we attempt to create sequences which contain one <math>2</math> and the rest <math>1s</math>, the sequence will contain one <math>2</math> and four <math>1s</math>. We can place the <math>2</math> in either the first, second, third, fourth, or fifth position, giving a total of five possibilities. If we attempt to create sequences which contain one <math>3</math> and the rest <math>1s</math>, the sequence will contain one <math>3</math> and three <math>1s</math>. We can place the <math>3</math> in either the first, second, third, or fourth position, giving a total of four possibilities. For sequences which contain exactly two <math>2s</math> and the rest <math>1s</math>, the sequence will contain two <math>2s</math> and two <math>1s</math>. The two <math>2s</math> could be next to each other, separated by one <math>1</math> in between, or separated by two <math>1s</math> in between. We can place the two <math>2s</math> next to each other in three ways, separated by one <math>1</math> in two ways, and separated by two <math>1s</math> in only one way. This gives us a total of six ways to create a sequence which contains two <math>2s</math> and two <math>1s</math>. Note that we cannot have a sequence of only <math>2s</math> and <math>3s</math> since the sum will either be <math>5</math> or greater than <math>6</math>. We now only need to consider the case where we use all three numbers in the sequence. Since all three numbers add to <math>6</math>, the number of permutations of the three numbers is <math>3!=6</math>. Adding up the number of sequences above, we get: <math>3+5+4+6+6=24</math>. Thus, answer choice <math>\boxed{\textbf{(E)}\ 24}</math> is correct.
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A dynamics programming approach is quick and easy. The number of ways to climb one stair is <math>1</math>. There are <math>2</math> ways to climb two stairs: <math>1</math>,<math>1</math> or <math>2</math>. For 3 stairs, there are <math>4</math> ways:  
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==Solution 2==
 
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A recursive approach is quick and easy. The number of ways to climb one stair is <math>1</math>. There are <math>2</math> ways to climb two stairs: <math>1</math>,<math>1</math> or <math>2</math>. For 3 stairs, there are <math>4</math> ways:  
 
 
(<math>1</math>,<math>1</math>,<math>1</math>)
 
(<math>1</math>,<math>1</math>,<math>1</math>)
 
(<math>1</math>,<math>2</math>)
 
(<math>1</math>,<math>2</math>)
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Thus, there are <math>\boxed{\textbf{(E) } 24}</math> ways to get to step <math>6.</math>
 
Thus, there are <math>\boxed{\textbf{(E) } 24}</math> ways to get to step <math>6.</math>
  
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==Solution 3==
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== Video Solution by OmegaLearn ==
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Complementary counting is also possible. Considering the six steps, Jo has to land on the last step, so there are <math>2^5=32</math> ways to land on the other five steps. After that, subtract the number of ways to climb the steps while taking a leap of <math>4</math>, <math>5</math>, or <math>6</math>. The eight possible ways for this is (<math>4</math>, <math>1</math>, <math>1</math>), (<math>4</math>, <math>2</math>), (<math>1</math>, <math>4</math>, <math>1</math>), (<math>1</math>, <math>1</math>, <math>4</math>), (<math>2</math>, <math>4</math>), (<math>1</math>, <math>5</math>), (<math>5</math>, <math>1</math>), and (<math>6</math>)
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https://youtu.be/5UojVH4Cqqs?t=4560
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~ pi_is_3.14
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==Video by MathTalks==
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https://youtu.be/mSCQzmfdX-g
  
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Altogether this makes for <math>32-8= \boxed{\textbf{(E) 24}}</math> valid ways for Jo to get to step 6. -adyj
 
  
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== Video Solution ==
 
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https://youtu.be/5UojVH4Cqqs?t=4560
 
  
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~ pi_is_3.14
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=24|after=Last Problem}}
 
{{AMC8 box|year=2010|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:12, 15 July 2024

Problem

Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$, $2$, or $3$ at a time. For example, Jo could climb $3$, then $1$, then $2$. In how many ways can Jo climb the stairs?

$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution 1

A dynamics programming approach is quick and easy. The number of ways to climb one stair is $1$. There are $2$ ways to climb two stairs: $1$,$1$ or $2$. For 3 stairs, there are $4$ ways: ($1$,$1$,$1$) ($1$,$2$) ($2$,$1$) ($3$)

For four stairs, consider what step they came from to land on the fourth stair. They could have hopped straight from the 1st, done a double from #2, or used a single step from #3. The ways to get to each of these steps are $1+2+4=7$ ways to get to step 4. The pattern can then be extended: $4$ steps: $1+2+4=7$ ways. $5$ steps: $2+4+7=13$ ways. $6$ steps: $4+7+13=24$ ways.

Thus, there are $\boxed{\textbf{(E) } 24}$ ways to get to step $6.$

Video Solution by OmegaLearn

https://youtu.be/5UojVH4Cqqs?t=4560

~ pi_is_3.14

Video by MathTalks

https://youtu.be/mSCQzmfdX-g



See Also

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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

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