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Exerciz 2 Let n be pasitive integer The Koal of this exercise is to give combinatorial proof for the identityFn+ =2() where Fn is the n-th Fibonacci number starting...

Question

Exerciz 2 Let n be pasitive integer The Koal of this exercise is to give combinatorial proof for the identityFn+ =2() where Fn is the n-th Fibonacci number starting with Fi = Fz = 1 Recall that, COM binatorial proof;, one uSually counts the same quantity in two different ways: suhset 5 € {1, n} is said to be slithy if all elements of 5 are geatu than the cardinality of 5_ For example the empty sct is slithy (), but {1} is not slithy:Let Sn denote the number of slithy subsets of {1= (i) Establi

Exerciz 2 Let n be pasitive integer The Koal of this exercise is to give combinatorial proof for the identity Fn+ = 2() where Fn is the n-th Fibonacci number starting with Fi = Fz = 1 Recall that, COM binatorial proof;, one uSually counts the same quantity in two different ways: suhset 5 € {1, n} is said to be slithy if all elements of 5 are geatu than the cardinality of 5_ For example the empty sct is slithy (), but {1} is not slithy:Let Sn denote the number of slithy subsets of {1= (i) Establish the recurrence relation sn+1 Sn-1: Deduce thuat Sn (quals the left-hand side By counting the slithy subsets of size k = 0.1, show that sn equals the right-hand side



Answers

Use a combinatorial argument to show that the sequence $\left\{D_{n}\right\},$ where $D_{n}$ denotes the number of derangements of $n$ objects, satisfies the recurrence relation
$$D_{n}=(n-1)\left(D_{n-1}+D_{n-2}\right)$$
for $n \geq 2 .[\text { Hint: Note that there are } n-1 \text { choices for the }$ first element $k$ of a derangement. Consider separately the derangements that start with $k$ that do and do not have 1
in the $k$ th position. $]$

Okay, so we want to prove that the follower for realize true, but checking that now this is the same the different ways off counting the same thing. So we want to think about these as all the ways we can select Gary elements and distinguish two of them that equal menacing. So let's let's think about it this way. So we have and bulls in a box, we want to sell it, kill them, and from those case, you're gonna choose your favorite, and I'm gonna choose my faith. So let's see how we do that. So we know. But each time we can select from the in balls, okay. And I have k options for my favorite, and you have K options for your favorite that it could be the same. Okay, but this is only for one choice. Um, how many ways come we choose this element? Well, could just sear of them. We could choose to You could choose one until then because that's the amount that we have. So this is all the ways that we can select some balls from the box and you choose your favorite and I trust mine off that selection Okay, so that's what these first sight means. Now, that's the second side means the same. Well, it's not that easy to see it on this expression, but it's gonna be somewhere is here to see it in the side. Now, I know this is a little bit hard. Later to explain on your homework, for example, try to write with words on what were how explaining. So now suppose that we're looking at the evidence. Okay, this'd is each one other. Um, and the first thing I'm gonna do is I'm gonna go ahead and select my favorite. Let's say my favorite is the last one. So I selected this and I say a Let's first I choose one and you to study for one. So I have and choices for my favorite. And since I don't want you to choose the same that I chose, you have in minus one possibilities for yours because you have one less. So now we selected our favorites, and we have in minus two left, and then we're gonna decide if we want them or not. So we're gonna be the one who wanted to be selective. And a seraph is now So for sure. We're selecting the ones that we like, and then we have randomly decided to choose or not choose the rest. So for each one of them have two options. We decide the boot seer or one, and we are assigning Ciro's or one for in minus two for the left over. So we have and minus two twos. Thank you. Best the first, though. Now I say, OK, seems we before decided to choose a different one. It's just the same one now. Okay, Money. A bit bossy. So I'm the one who's telling you what's true, sir. Okay, so we have an elements now, and we're both gonna choose the same one. So we decide that this one is a champion. So we had in options for that one because we're bitches. Any of those end you know, we have in minus one left toe, which we will assign one if we chose it. A Ciro if were the end. Okay, so for each one of them, we have two options and things that are in minus one of them. These in minus one Options. Okay, So what we have here is exactly the same. That we will happen in the heat. Okay? So for what we have talk about, we prove that these two things are the same. Now the question is, can I arrive? The last one I said us something related to this intentional Siri's. Yes. And it's not too bad because we can extract that in from both of them. And we can actually also extract to do that and minus two because we have in minus two here in minus one there so that it's more actually here on there. And so what we have left is from the 1st 1 in minus one him from the second. Then we have 12 ladies left, and that is actually in bless one. So we shot in thanks and plus one two. So the n minus two, proving that this identity is true

Indiscretion. We are asked to prove that the formula for the number off party shuns off the separate and elements. Is this formula when we are given the initial Cornish under pee off zero is one. We need this too. Like we're gonna use in the shun strong induction to prove this. And we need to show in basic step that p of one is correct, right to the formula. But then in pee off one, we will need p of zero two to show that the formula is correct and we have given that so everything is fine now, the strong induction I mentioned is a bit different than no more induction in that we as to not just not does some k and then show okay, plus one is true. We will assume that everything from 12 tree up to this point to care are all true. And then we gonna show care plus one. We can't do this. It is a legit program. It give the same result. It just require a bit more like condition that we set up. But since the formula were given is you sing everything before it so we are forced to use this strong induction instead, and less do this. So on the basic step, as I said, set off one element. There's only one condition right itself and the formula Give this which is one so basic step is clear now for inductive step. We assume that the formula is true for and equal to 12 tree up and you que for some positive vintage, Okay. And we want to show that it is true for kid plus one as well, you see can show this using the information of both. Then we will have prevent the statement. Okay, so now consider the scent off K plus one element. When we are looking at a partition, he's like the newest element. I will. I will thicken in a set off his own in every position we're looking at. And for the first for the first case, when we fix the last the newest element in a set. Everything else in the petition, like how many positions are there that has last element in one said alone? Well, the answer is p off k right, because what we have to do is used the K element that are left in the set. Create a partition off them. And how many other is p? Okay, which we assume by in the China help a disease that it exists. And like that pee off Kay is the number off Patricia, I should say, OK, that is the first case. Next we add one element in there. Which element we're gonna add. We can choose from the KLM in left Interstate. Right? So they're gonna be kids who's won way to issues element to feel the said and And what left gonna be K minus one? And those can be those who have the petition. Like, how many polish on is it? Pee off came in this one. You see the pattern we keep adding we keep adding element in this set with with last element in it now issues too. So it's gonna be K shoes too. Bye, Andrew. Whatever is left we No, that there are pee off came and as to way to mayor position all day and we continue and we're gonna and up with the last. The last case would be when everything else in the city So from a one a two up until a K all inside. It said, together with a kid plus one, and we and we have nothing left. So this is, in fact, they'll a PR zero. Okay, p of zero way to to make up tradition. And so we add these cases together, you're going to see the pattern that it is. In fact, the some off this this I last element would be last case will be K shoes care. Right. So here we from Katie shoes J J Iran from zero to kay Times issue ish kistane P off K minus d Right, Because here is p zero. Yeah, he is Day is Tuesday. Okay, Manners to Jay's once. Okay, minus one. And so here we have the formula for kid plus one, which is what we want from the start. So we have proven the inductive step. All right? And so to get the with basic step, we have proven the statement for any That is true for the send off in limine for any positive integer. Okay, that is it. Thank you


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