1 Step 2 Step

1019 /www/nrich/html/content/00/10/penta1/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Liam's house has a staircase with $12$ steps. He can go down the steps one at a time or two at a time. For example: He could go down $1$ step, then $1$ step, then $2$ steps, then $2$, $2$, $1$, $1$, $1$, $1$. <center><mdo:image alt="1 or 2?" src="penta1.gif"></mdo:image></center> In how many different ways can Liam go down the $12$ steps, taking one or two steps at a time? <a href="http://nrich.maths.org/7199">Click here for a poster of this problem</a>. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">This problem is a tricky one because it can be a bit tempting to try to count all the possibilities one by one. While this is sometimes necessary, often there's a better approach. Tayla, Cindy and Iris from Elm Park School started with smaller flights of stairs and spotted a pattern. Miss Hodgetts' Y9 Class from Ounsdale High School and many other people also saw the pattern. Well done. Suppose Liam is going down $12$ steps. He can start by taking one step, then he has $11$ more steps to go down. The only other possibility is that he starts by taking two steps and has $10$ more steps to go down. So how does the number of ways of going down $12$ steps depend on the number of ways of going down $10$ steps and $11$ steps? It helped to start with small numbers of steps and many people sent in lists like this one: <table border="1"> <tbody> <tr> <td>1 Step</td> <td>1</td> <td>Total 1 way</td> </tr> <tr> <td>2 Steps</td> <td>1,1 or 2</td> <td>Total 2 ways</td> </tr> <tr> <td>3 Steps</td> <td>1,1,1 or 1,2 or 2,1</td> <td>Total 3 ways</td> </tr> <tr> <td>4 Steps</td> <td>1,1,1,1 or 1,1,2 or 1,2,1 or 2,1,1 or 2,2</td> <td>Total 5 ways</td> </tr> <tr> <td>5 Steps</td> <td>1,1,1,1,1 or 1,1,1,2 or 1,1,2,1 or 1,2,1,1 or 1,2,2 or 2,2,1 or 2,1,2 or 2,1,1,1</td> <td>Total 8 ways</td> </tr> <tr> <td>6 Steps</td> <td>1,1,1,1,1,1 or 1,1,1,1,2 or 1,1,1,2,1 or 1,1,2,1,1 or 1,2,1,1,1 or 1,1,2,2 or 1,2,2,1 or 2,1,1,2 or 2,1,2,1 or 2,2,1,1 or 2,2,2 or 2,1,1,1,1 or 1,2,1,2</td> <td>Total 13 ways</td> </tr> <tr> <td>etc.</td> <td> </td> <td> </td> </tr> </tbody> </table> The tricky pattern to spot is that each total on the right is the sum of the previous two totals. Peter Zhu pointed out that this can be expressed simply as an algerbaic relation. He used the notation $S_n$ to mean the total number of ways of descending $n$ steps. Then the formula is $S_n = S_{n-1} + S_{n-2}$ So, since $S_1=1$ and $S_2=2$, that means $S_3=3$, $S_4=5$, $S_5=8$, $S_6=13$, $S_7=21$, $S_8=34$, $S_9=55$, $S_{10}=89$, $S_{11}=144$, and $S_{12}=233$. The number of ways to get to the 12th step is 233. The reason for this pattern is very simple. For the first step Liam has two choices: i) Jump down 1 step. This leaves n-1 steps. ii) Jump down 2 steps. This leaves n-2 steps. So if you know that Liam has $S_{n-1}$ ways of going down n-1 steps, and also $S_{n-2}$ ways of going down n-2 steps, all of these, when combined with the right choice of first step, give a way of going down n steps. This sum is exactly the relation $S_n = S_{n-1} + S_{n-2}$. Congratulations to everyone who managed to spot this pattern and find this answer. Those of you who have heard of the Fibonacci numbers will notice that the totals are the Fibonacci sequence, starting at 1, 2, ... In fact there is a formula for finding any Fibonnaci number without having to add up all the previous ones. You might like to try and find out what it is. This would make counting the total number of ways of jumping down 1000000 steps much easier, but you'll probably never need to jump down that many steps. Another interesting detail is that there are many number sequences that follow the pattern $S_n = S_{n-1} + S_{n-2}$. The Lucas numbers are an example. Can you see that if you specify the first two numbers, all the rest of the sequence is given without any more choice? What about if you are told only the first number, how many possible sequences with this addition property are there? There is another reasonably quick and easy way to find the answer 233 if you're confident with combinations and binomial coefficients. These are some stage 5 topics that you can read about in <a href="http://nrich.maths.org/7713">this article</a>. Matias Cano sent in his idea: First of all I broke down all the different ways of down 12 stairs: <ol> <li>Always jump one step at a time</li> <li>Jump two steps 1 time, and one step 10 times</li> <li>Jump two steps 2 times, and one step 8 times</li> <li>Jump two steps 3 times, and one step 6 times</li> <li>Jump two steps 4 times, and one step 4 times</li> <li>Jump two steps 5 times, and one step 2 times</li> <li>Always jump two steps at a time</li> </ol> Matias then used binomial coefficients. These strange and useful numbers have a special notation and formula, explained in more detail in the <a href="http://nrich.maths.org/7713">article</a>. $n \choose r$ means 'the number of ways of choosing $r$ objects from a total of $n$' where the order of the choices doesn't matter. It's said 'n choose r' For example ${4 \choose 2} = 6$ because from the numbers 1, 2, 3, 4 I can make six pairs: 12, 13, 14, 23, 24, 34 (remember we consider 34 the same as 43) The formula for the binomial coefficients uses the factorial notation $n! = n\times(n-1)\times(n-2)\times...\times 2 \times 1$ and is $${n \choose r} = \frac{n!}{r!(n-r)!}$$ So the factorials on the right give a formula for 'n choose r'. This can be linked to jumping down stairs because Liam can decide where to put his two-step jumps. If there are n total jumps, and exactly r of them are two-step, then there are $n \choose r$ possible ways of making the jumps. We have to 'choose' r different places out of the n steps to put the double jumps. For the item 1 there's 1 posibility For the item 2 there are 11 posibilities because this is 11 choose 1 For the item 3 there are 45 posibilities because this is 10 choose 2 For the item 4 there are 84 posibilities because this is 9 choose 3 For the item 5 there are 70 posibilities because this is 8 choose 4 For the item 6 there are 21 posibilities because this is 7 choose 5 For the item 7 there are 1 posibility Total of the posibilities will be: 1+11+45+84+70+21+1 = 233 Final result = 233 posibilities. Brilliant work everyone. This was a problem which can involve some sophisticated and difficult maths! </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem ?</h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=1019">This problem</a> provides an opportunity to draw out from students techniques for problem solving.</div> <div>A key to solving this problem is to simplify to smaller staircases, and use these earlier results explain what is happening and hence calculate later results.</div> <h3>Possible approach</h3> The Article <a href="/2338">Go Forth and Generalise</a> may be of interest. "Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at a time. For example: He could go down 1 step, then 1 step, then 2 steps, then 2, 2, 1, 1, 1, 1. In how many different ways can Liam go down the 12 steps, taking one or two steps at a time?" Give students some time to explore the problem. As they work, wander round the classroom, listening for any particularly useful insights. In particular, listen out for comments like: "There are going to be loads of different ways" "How are we going to be able to make sure we don't miss any?" Then bring the group together to share those insights. If no-one has suggested it: "Perhaps we could work on a simpler version of the problem to see if that helps? Let's see how many ways there are of going down three, four, five or six steps." Students could all work on these or they could be shared out with different groups working on different sizes of staircase. Then bring the class together and invite students out to the board to list the number of different ways they found, while the rest of the class make sure they have worked in a way that makes sure they haven't missed any. Once the number of ways for three, four, five and six steps have been worked out: "With your partner, look at the number of ways for each different size of staircase, and see if you notice any patterns. In a short while we will return to the original problem of a staircase with twelve steps. Can you use what we have found out about smaller staircases to make predictions about the answer for twelve steps? Or any number of steps?" Give students some time to work with their partner, and circulate, listening in for any who make connections. If they are struggling to make progress, here are some prompts you could use: "How does the number of ways of descending 5 steps compare to the number of ways of descending 3 and 4 steps? What about the number of ways of descending 6 steps compared to the number of ways of descending 4 and 5 steps?" "If I want to go down 5 steps and I start with a one-step, how many ways can I descend the remaining four steps?" "If I want to go down 5 steps and I start with a two-step, how many ways can I descend the remaining three steps?" Finally, bring the class together and discuss their findings. They could present their answer to the twelve-step problem or an explanation of the general case on a poster. <h3>Key questions</h3> <div>What is making it difficult?</div> <div>What have you found out so far?</div> <div>Key comments students may make:-</div> <div>"There'll be too many", "I can't keep track", "I might have done some twice"</div> <div>Possible responses:-</div> <div>"Work it out for fewer steps", "Try to find a logical way to order the options", "Work together and check each other's work".</div> <div>Students will generate lots of numbers and have to find a way to keep track of what they've done, ensuring that each set of results is complete</div> <h3>Possible extension</h3> For more challenging tasks that involve working systematically, see this <a href="/9545">collection</a> for Upper Secondary students. <h3>Possible support</h3> <div>Begin with a smaller number of steps, so it is still too big, but not quite so daunting.</div> <div>Another problem that uses a similar idea is <a href="http://nrich.maths.org/public/viewer.php?obj_id=4338&part=">Colour Building</a></div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Making careful lists, counting and looking for a pattern are useful strategies for exploring this problem. Once you spot a pattern you should try to work out why the pattern occurs. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Suppose Liam is going down $12$ steps. He can start by taking one step, then he has $11$ more steps to go down. The only other possibility is that he starts by taking two steps and has $10$ more steps to go down. So how does the number of ways of going down $12$ steps depend on the number of ways of going down $10$ steps and $11$ steps? Making careful lists, counting and looking for a pattern were useful strategies for solving this problem, which is nicely explained by Christopher fromBurpham Primary School, Guildford, Surrey: "Following the hint to try one step, then two and so on, I listed all the combinations of answers for the first few numbers of steps. Once I had got to 7, I thought about finding a pattern to help me work out the bigger numbers. I noticed that the next number in the sequence was the sum of the two previous numbers. This meant I could work out the answer, which is 233. " Alex from The Abbey School, Woodbridge, Suffolk sent in a well made list of all the possibilities up to 8 steps, as well as finding the pattern: <table border="1"> <tbody> <tr> <td>1 Step</td> <td>1</td> <td>Total 1 way</td> </tr> <tr> <td>2 Steps</td> <td>1,1 or 2</td> <td>Total 2 ways</td> </tr> <tr> <td>3 Steps</td> <td>1,1,1 or 1,2 or 2,1</td> <td>Total 3 ways</td> </tr> <tr> <td>4 Steps</td> <td>1,1,1,1 or 1,1,2 or 1,2,1 or 2,1,1 or 2,2</td> <td>Total 5 ways</td> </tr> <tr> <td>5 Steps</td> <td>1,1,1,1,1 or 1,1,1,2 or 1,1,2,1 or 1,2,1,1 or 1,2,2 or 2,2,1 or 2,1,2 or 2,1,1,1</td> <td>Total 8 ways</td> </tr> <tr> <td>6 Steps</td> <td>1,1,1,1,1,1 or 1,1,1,1,2 or 1,1,1,2,1 or 1,1,2,1,1 or 1,2,1,1,1 or 1,1,2,2 or 1,2,2,1 or 2,1,1,2 or 2,1,2,1 or 2,2,1,1 or 2,2,2 or 2,1,1,1,1 or 1,2,1,2</td> <td>Total 13 ways</td> </tr> <tr> <td>7 Steps</td> <td>1,1,1,1,1,1,1 or 1,1,1,1,1,2 or 1,1,1,1,2,1 or 1,1,1,2,1,1 or 1,1,2,1,1,1 or 1,2,1,1,1,1 or 1,1,1,2,2 or 1,1,2,2,1 or 1,2,2,1,1 or 1,1,2,1,2 or 1,2,1,1,2 or 2,1,1,1,2 or 2,1,1,2,1 or 2,1,2,1,1 or,2,2,1,1,1 or 2,2,2,1 or 2,2,1,2 or 2,1,2,2 or 1,2,2,2 or 2,1,1,1,1,1 or 1,2,1,2,1</td> <td>Total 21 ways</td> </tr> <tr> <td>8 Steps</td> <td>1,1,1,1,1,1,1,1 or 1,1,1,1,1,1,2 or 1,1,1,1,1,2,1 or 1,1,1,1,2,1,1 or 1,1,1,2,1,1,1 or 1,1,2,1,1,1,1 or 1,2,1,1,1,1,1 or 2,1,1,1,1,1,1 or 1,1,1,1,2,2 or 1,1,1,2,1,2 or 1,1,2,1,1,2 or 1,2,1,1,1,2 or 2,1,1,1,1,2 or 1,1,1,2,2,1 or 1,1,2,2,1,1 or 1,2,2,1,1,1 or 2,2,1,1,1,1 or,1,1,2,2,2 or 1,2,1,2,2 or 2,1,1,2,2 or 1,2,2,1,2 or 2,2,1,1,2 or 1,2,2,2,1 or 2,2,2,1,1 or 2,2,2,2 or 1,2,1,2,1,1 or2,1,2,1,1,1 or 2,1,1,2,1,1 or 1,1,2,1,2,1 or 1,2,1,1,2,1 or 2,1,1,1,2,1 or 2,1,2,2,1 or 2,2,1,2,1 or 2,1,2,1,2</td> <td>Total 34 ways</td> </tr> <tr> <td>etc.</td> <td> </td> <td> </td> </tr> </tbody> </table> I noticed that the number you are trying to get is always the sum of the 2 numbers before it i.e. the Fibonacci series, so I think the next ones will be: 9 Steps:55 ways (34+21) 10 Steps:89 ways (55+34) 11 Steps:144 ways (89+55) 12 Steps:233 ways (144+89) Excellent solutions were also sent in by: Olivia from St Ives School, Haslemere; Melis Onalan , Umur Ta Demir and Altay Alpgut from Private IRMAK Primary School, Istanbul, Turkey; Daniel from Anglo-Chinese School, Singapore; and Year 6 Maths Club, St Francis School, Maldon, Essex. This last solution has been chosen because it tells the story of a team effort in finding a good solution. It is from: Emily, Caroline, Rebecca, Clare, Camilla and Gabriella from The Mount School, York. "We started by drawing diagrams, like this <mdo:image alt="" src="Image1.gif"></mdo:image> But Emily had the idea of just using the numbers, so <table border="1"> <tbody> <tr> <td> <div>For 2 steps</div> </td> <td> <div>1,1 & 2</div> </td> <td>2</td> </tr> <tr> <td> <div>For 3 steps</div> </td> <td> <div>1,1,1 & 1,2 & 2,1</div> </td> <td>3</td> </tr> <tr> <td> <div>For 4 steps</div> </td> <td> <div>1,1,1,1 & 2,1,1 & 1,2,1 & 1,1,2 & 2,2</div> </td> <td>5</td> </tr> <tr> <td> <div>For 5 steps</div> </td> <td> <div>1,1,1,1,1 & 1,2,1,1 & 1,1,2,1 & 1,1,1,2 & 2,1,1,1 & 2,2,1 & 2,1,2 & 1,2,2</div> </td> <td>8</td> </tr> <tr> <td>For 6 steps</td> <td> <div>1,1,1,1,1,1 & 2,1,1,1,1 & 1,2,1,1,1 & 1,1,2,1,1 & 1,1,1,2,1 & 1,1,1,1,2 & 2,2,1,1 & 2,1,2,1 & 2,1,1,2 & 1,2,2,1 & 1,2,1,2 & 1,1,2,2 & 2,2,2</div> </td> <td>13</td> </tr> </tbody> </table> At this point Camilla came up with 'Fibonacci' and the cry was taken up by the others ! 2 +3 = 5 3 + 5 = 8 5 + 8 = 13 so ..... 21, 34, 55, 89, 144, 233 The answer is 233" Here is another way of looking at the problem which may make it easier to see why the Fibonacci series occurs: Liam's staircase has 12 steps, and he must start EITHER taking just a single step OR by taking a double step. If he takes a single step to start with, then he has 11 steps left. If instead he takes a double step to start with, he only has 10 steps left. So the number of ways he can go down all 12 steps is equal to the the number of ways to go down 11 steps PLUS the number of ways to go down 10 steps. But that's just the same as the way the Fibonacci numbers are defined - each number is the sum of the two numbers that came before it. So the answer is just the 12th Fibonacci number! We could use some simple notation to help explain what we mean: Let's call the number of ways that Liam can go down a staircase with n steps $S_n$. (What we eventually want to know is $S_{12}$, the number of ways he can go down his 12-step staircase.) Liam must start off taking EITHER a single step OR a double step. If he takes a single step to start with, then he has $n-1$ steps left and there are $S_{n-1}$ ways he could do those. If instead he takes a double step to start with, he only has $n-2$ steps left and there are $S_{n-2}$ ways he could do those. So the total number of ways he can do all n steps is the sum of these two: $$S_n = S_{n-1} + S_{n-2}.$$ But that's just the definition of the Fibonacci numbers! So the answer is the 12th Fibonacci number. Vicky from Cambridge pointed out: <div>Though the pattern from walking down the stairs satisfies the recurrence relation $S_n = S_{n-1} + S_{n+2}$ as shown above, there are many sequences that satisfy this, including the Fibonacci and Lucas numbers. To clarify which sequence it is, you must define the first two terms.</div> <div> </div> <div>In fact the Fibonacci sequence is defined with $F_0 = 0, F_1 = 1$ and so on.</div> <div>So in fact the answer for going down $12$ steps is in fact the $14^{th}$ Fibonacci number.</div> </mdoxml> 2 3 0 0 1 0 0 1 Step 2 Step Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps? Numbers and the Number System Integers Sequences, Functions and Graphs Sequences Using, Applying and Reasoning about Mathematics Mathematical reasoning & proof Handling, Processing and Representing Data Processing and representing data Decision Mathematics and Combinatorics Combinations Sequences, Functions and Graphs Fibonacci sequence Information and Communications Technology smartphone Collections Stage 3&4 Investigation Secondary Mapping Document MD Patterns and sequences US Secondary Mapping Document DisplayCabinet Secondary processes PM - Exploring and noticing structure Secondary processes PM - Working Systematically ajk44 solution needs editing Secondary processes PM - Reasoning, Justifying. Convincing. Proof.