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Use the pigeonhole principle to solve the following problem. LetX = {0, 1, 2, 3, 4, 5, 6}. Show that for any choice of 17 subsetsof X with cardinality at most 3, at...

Question

Use the pigeonhole principle to solve the following problem. LetX = {0, 1, 2, 3, 4, 5, 6}. Show that for any choice of 17 subsetsof X with cardinality at most 3, at least two such subsets are suchthat their elements sum up to the same number. For instance, thesubsets {1, 2, 6} and {0, 4, 5} are such that the sum of theirelements is 9.

Use the pigeonhole principle to solve the following problem. Let X = {0, 1, 2, 3, 4, 5, 6}. Show that for any choice of 17 subsets of X with cardinality at most 3, at least two such subsets are such that their elements sum up to the same number. For instance, the subsets {1, 2, 6} and {0, 4, 5} are such that the sum of their elements is 9.



Answers

Suppose that six distinct integers are selected from the set $\{1,2,3,4,5,6,7,8,9,10\} .$ Prove that at least two of the six have a sum equal to $11 .$ Hint: Consider the partition $\{1,10\},$ $\{2,9\},\{3,8\},\{4,7\},\{5,6\}$.

In this problem were asked to find the number of subsets that could be created out of this set. This set has 10 elements in it. One through 10. Um, so a subset would be any grouping of any elements of that set so you could have no elements have been empty set. You could have all the same elements. That would be all 10 elements or any other combinations. You could have just the number one, just the number three, you know, the one end of three any combination. So the formula to do this is U to the power of end or end. It's the number of elements. So in this case, we have to the power of 10 which equals 1000 and 24.

Okay. So for this exercise we need to use a tree diagram to determine the number of subsets of this set formed by 39 11 and 24. Such that the sum of the elements is less than 28. Okay, so let's start with the initial note. It's going to be the empty set and then we are going to have the singletons formed by each of the elements of this set. Okay, so that rated better. So here we are going to have the element 24 here. 11. Here to nine. Here, Deep, I'm here. I'm missing one element is the seven the nine marriage. This 4 11 nine seven three. Okay, so here each connection is going to form a subset. Okay, So in this case we're considering the single towns. Okay, that means the signal 24 11 97 and three. No, until here all these subsets are the some of these subsets are less than 28. Right now, we can add to this set of 20 for one more element in a way that the sum is still less than 28. Yes, the element that we can add is the number three. So at the end of this, at the root of this, this this branch of the the three we have here to set 3 24. Okay. Is like the each connection means a union. Okay, so the empty set union for here we have a single in turn set T V p the 24 that only contains 24. And here we we joined the 24 with the number three. You only have this subset. So until here we have 20. Less than 28. Okay. We cannot add any other element to this branch because otherwise we're going to exceed the 28. Now, let's see what happened with the 11. So to the 11 we can add the number um nine because that is going to some 20 we can add the number seven because that is going to be 18. And we can add of course the number three because that is going to be 14. And we can continue of course. So here until until this point we are some in 20. Okay, so we can add the number three because we're going to have 23 we can add of course to this upset the number seven Until that we're going to have 27. That is less than 28. So it's correct here we have the seven. Okay, joined with the 11. So we have 18 and we can act here just the number three because we don't have any other elements. The number nine is already considered in this part of this branch. Okay, so seven and three and 43. We cannot put any other element because those subsets are already considered in these branches here and here. Okay, so that's you know what happened here with the nine? Well, we are not going to at the 11 because that is already considered, but we can add the element seven here. So let me move a little bit this here. So to the nine, we can add the number three. So we have until this point a subset and the sum is going to be 12. So respect the condition and we can add also the element seven. Why can we not at the 11? Because that subset is already considered here. So it's not necessary to adhere the level but 97 is not considered. Okay. And that's upset. It's going to be less than 28. And of course from here we can add the element three and that's going to be 19. And again, he was upset that satisfy the conditions now to the seven here, you can observe that the seven has appear here here. So the only element where the seven is not joined to is the element three. I mean just these two elements and not because seven with 7, 11 is already considered in this part. Seven with nine is considered in this part here. And seven with 24 is impossible because that is going to exceed 28. So the set of the only element that is not joined with seven is three and three here is going to left along because all the possible combinations that of elements that can join with the number three are already concept. And this three give us the oldest subsets that some elements that some of the elements is going to be less than 28. The Big Blue

It is given that number off elements in the set are seven. We need to find the number off subsets. Now, since we know that the number of subsets off are set containing an element is equal to to restore power. And so, since our set contains seven elements so number off subset is equal toe tourist to power seven, which is equal toe 1 20 it so this is a real answer.

In this question, we are asked to prove that the set with an element when Aeneas at least tree has end times in minus one times in minus two over six subset. This is the number of sulfites contending exactly three elements. So there will be something. Looks like this may be any any substance that has treated element in it. Who will be count on DDE that we claim that this is the formula We will prove you're seeing induction as always Include when include in like this is always induction and basic step We started an equal to tree We have exactly one substance, right? And the fibula also give number one So basic step is clear Now next we look at inactive state even suppose Ah sorry this whole Okay This host that meant is pure when we suppose p o. N is true for in greater than or equal to tree some in we will show that then the next one The peon plus one is also true We have to use peon as an evil mission somehow And let's do this first from p m every have this They want to it Have a sit with an element. Then we have this many subset already established. What happened? What happened If we add one more and the new element in there to make the total number in plus one, what happened is we will look at the the news upset created right from from this phenomenon. And how do we call them? Well, the O's upset is the old subsets are considered all Onley element like that already exist there. The news have said would have to have X in plus one. Right, So all of these upset already has this one. What about it? The last two slot. Well, it can be anything from from the the already established said so from a one to it in anything can be in there. So the number off configuration from this is in shows too, right? Because we can choose any two women in India set to feel this role and we can choose exactly inches two way on some some people who is and see to Yeah, it's the same thing. It has a formula and its former light years. This is in is actually in fact Oh, uh, too fat times in minus two fact, but it can It can be reduced to this. Okay. And now we add, we add this many subset into the total number what we would get. Well, we add them true. And we get this formula So we will have this menace upset now in the in, plus one element. And who is the P in plus one? Right? Because PM is this term pee and has won me shift into invest one So the middle will be humping the last will be in minus one and accounted for. So we have a show that the inductive step also works. Now we have proven both step off induction. So come back here. We have proven the statement and that is how you do it. Thank you.


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