That's deceptively simple, I think. No, I find that when a lot of questions when you actually can reduce the formula they are very straightforward but you know questions that are don't really quite you fold any categories or don't use formulas sometimes they are more tricky. Yeah, like when you see when I see it I go, Oh, okay. Yeah, but until you do, there's no, there's no, you know, there's no. Yeah, there's a machine and couldn't do the problem. Yeah, exactly. So like that's what math can glue is where like a lot of questions and oh you can reduce the standard formulas permutation combinations and you're done in like a few, you know, lies but there's some question I find that even the numbers are small, and the question numbers, maybe three or four points, but they can be quite tricky. Those are more like it's not about the math you know often before they give these just you know pattern recognition mathematical vision questions. Yes, just what this is. Exactly. Yeah, problem. Yeah, because if you already have formulas and you know how to use it but you are given something unknown you have to find out the pattern the way to do it. That's that's why I think that's kind of fun. makes it different. Feel free to work on this while we await the coming tide. Yeah, mommy of your students. So we have. You might think that maybe these two symmetry would have an even number, but actually isn't the answer is an odd number. Because you're, you're not allowed to start from why you have to start from x and that asymmetry is enough. That's a very good point though. Yeah, yeah. Not as symmetric as it looks. Yes, there's a start and an end. Yeah, so if we, I mean we definitely have to start from here right because actually to know we know where to go here so actually can start from here and we. This is we go to why so we this is a start and this is finished. And we can forget about x and y but even from here. It could be quite confusing. So actually the answer is five. You know, we can So you go here, go here and travels here, or go here, go here, go here, but after you do it for why you actually realize that there's something very special about the, the center cell, as it should be right if you come back to the center so you can you can enter the center cell from any cells. And it also from the center cells you also connect to the starting and finish. So, one way to actually do that is like, would look at the order in which you enter the center cell so you enter it first right you go here, and you go like this. If you enter it first, but you enter it. The second, then you go here first and you're here, but you mean from here you can always complete the loop. So you can start with from as a first one, or the second one. Third one. The fourth one. And the fifth one. So you can. The last one you go all the way here and you visit it. The last so actually this this for this five past. Does it make sense to everyone. So you can you can enter it and then you go to every other cells or you go to this first one. And then you enter it. That means the center cell is going to be the second step, or you go like here and hit the center cell will be the first step, you go all the way here enters the center cell go here exists so that'd be the first step or you can go here comes here and exit so the center says when we the fifth step. So actually you have a five different ways of going from s to f traversing each cell exactly once. Yeah, so I think it is a pretty good questions for beginning of, you know, four points part. Okay. So let's get started for our class today so do we do we do. Could be notorious but it's basically means the art of counting. Okay. And so we want to go through the basic, which is permutation and combination risk permutations is number of permutation of an object that we arrange them in a row. And then if you don't want to permute permuted or an object we just want to take out our objects, then the first one we have n ways and way to choose the first one, and minus one for the second one also way and this is n minus r plus one, because this is our term. So, when we multiply these together we want to use a sort of a nice small compact formula which is n factorial divided by n minus r Victoria and this guy's just n Victoria by zero Victoria, which is n Victoria. Yeah, I think if you, I hope everyone is clear about this and we're going to have a problems with this so we don't need to go through this example. Then, one more thing we want to be careful this when, when the elements are different right so this is permutation we repeated elements what it in those n elements, some of them are the same. Then we know that there are n factorial ways to permute n objects, and there's k factorial ways to permute those identical, I mean those k objects, but because they are identical, then we have to divide n Victoria by k Victoria. It's just like if you have a plane and excavator or truck, and two identical cars and in principle you have a five Victoria ways of arranging five toys, but because the two cars identical is then you have to divide by two Victoria. And, and these are some useful formulas. Sometimes I mean it's kind of just plug in but it's good to remember some of these one zero object and n objects the same only one way, taking one or taking n minus one is n and you know these are the general formula. Then combination in combination. We do not care about the, the order right so if you want to find the combinations of want to take our objects out and object, but we do not care about the orders, then, at first, we take, and we have a, this actually is equal to PNR right so it's like you permitted permitted and object out and objects, but then we divide by our Victoria, because we do not care about the order and there are Victoria different ways to arrange those things we think of it as a committee right so if you have a, if you take three people ABC, you really do not care who is who just as long as it takes three, but in combination, you would have ABC and ACB, you know, things like that so you get the idea so we have to define divided by and take to our Victoria when our is the number of objects. So, we're going to go through. I think these are really basic formulas and we would have some questions that are pretty sure that we go through all of these cases so I don't think we need to do the examples over here. Yeah, these are important. What are the basic techniques of counting case work right so a lot of time we divide a counting problem into exclusive cases the cases. I mean, ideally should not overlap, and then should cover the whole space right so we can we do not meet any cases or double count any cases. And sometimes when you have overlapping, then you have to, you know, compensate for that. Just like you have a Venn diagram and there's, you know, an overlap in here so we have to be careful about that. And then complimentary counting sometime we want to count, you know, this set but it's very difficult to cut it directly, is that Then we're going to call the complimentary set right and then we take the total and subtract this complimentary set and we get what we want. For example, if the question is how many three digit numbers are not divisible by 11. Right, but we know that it's easy to count the total of two digit numbers and it easier to count how many three digit numbers are multiples of 11 so I will take the indirect route it's easier, it's faster. And that's called complimentary counting. And then another counting techniques constructive counting sometimes we need to explicitly construct the items and how count how many ways we can do it. We have a, we have a very nice problem that uses technique constructive counting in today's class. And a lot of the time, this is a hard part of counting with restrictions. For example, you want to arrange seven Kids in a row, but maybe two kids want to stay next to each other or do not want to stay with each other. Then when the counting with restrictions, it's often easier to deal with the restrictions first. This is a subject that I don't think I see a lot in math kangaroo, but I think it's important one and usually sort of related to counting because when we can count all different cases, we can calculate the probability of something happens. And this applies to the case when all the outcomes of the event are equally likely, then we can just take the number of successful outcomes divided by the total number of possible outcomes to get the probability of that event happening. So, for example, if a face a fair six sided dice right has six faces with the six numbers, then what's the probability that when it is row to the number of dots on the face that is obvious and odd numbers and, you know, it's easy right just take one three by three cases divided by six, we get 0.5 as a probability Okay, so much for the theory, let's get started on problem solving. The first one in the equation, and times u times m plus b plus e plus r equal to 33. Each letter stands for a different digit. How many different ways are there to choose the values of the letters. So it's pretty self explanatory. These are the letters are different. So the digits are different. And we also build on what we did earlier, right? Using number theories or whatever tools we had earlier. I mean, for counting for some students, it could be very easy, and for some students, it could be very difficult. It depends on the experience, because this is not a topic that is usually covered in school, right, David? It also depends on... I think some people are just innately better at this than others. Counting was always my most difficult... I mean, I learned it, you know, I learned it eventually. I had to practice, it wasn't easy for me. Yeah, no, no, it was definitely not like when I see algebra, I say, oh, I can definitely solve it, you know. What I will say, though, is that I'm taking a real university course in math at my school. Yes. And we're learning this stuff, I'm not even kidding. Oh, sure. At a university course, so what I'm saying is, really, it's, yeah, it's not something you learn at school at all. No, it's not. So that's why I think we... I didn't really want to delve too much into theory, but I think we're going to figure it out. And I really want to cover a lot of questions today so that we actually... because there's no way we can cover all this in an hour. We spend time talking about theories, so I go pretty quickly, but I really want to hope we can do like seven, eight questions today. So let's do it together. 33, the first thing we need to factorize it, right, 3 times 11. And these are big, right? So I, you know, with three, four different... these are four different digits. There's no way that could be three. Then I know that three has to be n times u, right? So n times u equals to three, n times u equal to three, and m plus b plus e plus r equal to 11, right? That is for sure. And now this one, I know there's two ways of doing so, because n, and you could be, you know, this is the only way you can factorize it, n times three. And n times three, right? And then for this one, this is, we have to choose the 11. It could be 11, then with the sum of four, sum of four different digits, right? So what could it be? 11, what is it? It could be four, four plus two, I think. Six plus zero plus one. Four, so four zero. Okay, 5, 3, 2, 1, 11 is, it could be 5 plus 3 plus 2 plus 1, but this is no, right, because we already used 1 and 3 over there, so I think I have to increase this one, it's like 5, 5, I have 6 plus, sorry about it, I thought 6 plus 1 plus 0, 6 plus 2, okay, 6 plus 4, 6, 6 plus. There's a 0 too. Yes, there's a 0, definitely we need a 0 because we don't need 1 and 3, so it's a three number, so it's going to be 2 and 4 and 5, I'm sorry about that, 2 plus 4 plus 5. So what you actually notice here is that. Yes, so because that's how much they're different anyway, I mean over here, actually you don't really, just want to make sure, but because we know four different digits, it has to be, right, so I mean. These are the four small digits you have and they sum up to 11. Yeah, so you actually just get away, just immediately go 4 factorial, I mean you want to spend time to make sure, but you know that this is the case, you have them, they tell you so, so you can say 4 factorial because these are four different digits, so in this case we recognize that it's per mutation, right, so it's 4 factorial and these are, the choice is independent, so we can just multiply them and that's why we have a 2 times 24, which is 48, and you got it, so it's great, thank you. And this is 84. Yeah, I think the only way to learn it is just like go, and I promise that we have every single problem for every category that we described in the introductions. So that is a permutation of different items, items that are different. And now how many five different positive numbers have a product of their digits equal to a thousand? So in this case, again, you need to factorize it, but after you factorize it, what you can do? 2 cubed times 5 cubed, that's a factorization. And we have a 65 inches. That means they have to be smaller than nine. Okay, so what we have in this case, I think we can do it together, and then in the last step, you can, you know, while we are doing it, you can enter the answer. So we definitely 555, they cannot, you know, multiply together or whatever, so the 555, they stand on their own, right? And we have eight, so we could, we need two more digits, so it could be one and eight, or five, five, five, and two and four. Yeah, I think that's, that's pretty much it. I hope everyone gets to this point, right, this is like basic number theory, and then from here to the number of digits is the tricky point, because we, it's a permutation, right, permutation, because we are trying to arrange them, and we don't have a zero here, so we are good, any number can be the first digit permutation, but unlike the previous questions, here we have a repeated element, these five are the same, right, so permutation with repeated elements. Okay, good, and so what would be the answer? We have a definitely five factorial, because we have five of them, and remember, how many different ways we can permutate these five digits? Three factorial ways, right, so that basically gives me five times four, which is 20, right, and same over here, it actually, we use the same formula, two and four are different, and these three fives are the same, so we have another 20, so the answer would be 40. Okay, yeah, and it's good, we already got it, with our small statistic, yes, and so remember, permutation with different elements and permutation with repeated elements, that's two things, two separate things, okay, next one. Okay, so we change here and do a little bit of, you know, geometry over here, the horizontal and vertical distances between the points on the grid are equal to one centimeter, how many line segments with ends at the points in the grid are there that have length equal to five centimeters? So, Okay, so I got the first answer is C. So this one, definitely you want to use casework, right? It's a question to illustrate the method of counting using casework. We break down different cases. And I know a lot of, I mean, I already got answer with 12 because if you do it vertically, like you count it horizontally, you got 6, right? So, and then you do it vertically, you have another 6, so that's 6 plus 12. But what about, you know, the diagonal? Is there any other way you can, you can get 5 other than, you know, the straight, the horizontal or vertical lines? This is honestly more of a geometry track. Yeah, geometry, yeah, but, but, you know, it's like counting, but just say that, you know, a lot of time you see counting, it's just not those formula. Because in, actually, in Math.Chem.Group, you have a lot of, as you can see, you know, counting that like this. So doing, doing casework. So just, just, just a hint on how to make 5. You remember the triplet, Pythagorean triplets, right? 3 and 4 also give us 5. And in this case, I think that's the only thing that gives us 5. So I give you maybe another one. It's crazy to casework. Just go, you know, you have 5 and you have 5 in one direction, 0 in the other. That obviously works. 4 and 1 is not good enough. And then 4 and 1 is too small, but then 5 and 1 is too big. And you can just casework it and you will find, yeah. Horizontal, you have a 6, right? And then vertical, this is a straight horizontal, you know, this horizontal line, yes. Then we have 6 and then vertical, also 6. If you have any ideas, just, you know, send it to the chat. Let's just give maybe 30 seconds to see if we can find out. So these are the obvious, but you see, this is also question number 27, so it should not be that obvious. OK, great. I got the correct answer, 36. OK, so why do we get 36? Yes, that's correct. 36 because we have a 3 here, right? And 4 here. So just draw it and you have a 5 over here. I'm sorry. So 5. But if we move this triangle one position to the right, so we have two positions and we also move it two positions down. So 2 times 3, so we have a 6 in this configuration, right? And then if we look to this right, so we have in this way, this configuration, we have a 6. And then in this configuration, we have another 6. And then that is left and right and up and down as well. We have this configuration and they are all different, right? Because one side is 3, one side is 4. So this one as well. So we have a 6, 6, 6. That's 24. And 24 plus 12 is 36. That's correct. And casework and we know that they are not overlapping. So we are in good shape. Okay, next one. David, could you please do this question? How many 8-digit numbers, which consists, okay, A1, A2, A3 to A8, which consists of zeros or ones only. So A1 has to be 1, otherwise it's a 7-digit number, have the probability of the property that A1 plus A3 plus A5 plus A7 equals A2 plus A4 plus A6 plus A8. Which is most definitely a casework problem. Yeah, no, this is a really notorious question. A lot of times it's hard, like how, how did I divide into cases, right? Like how to count. Yeah, so this is a typical one that requires quite a bit of thinking. You can play a little bit with it. I mean, the numbers are small, so we can. In fact, they're as tiny as they could possibly be. It's 0 and 1. Yeah, but still we have to divide into cases. I mean, the number of digits is only 8, right, so we can do it one by one, but just play around with it a little bit. Yeah, this is a hard question. David, maybe you could... Yeah, I'm beginning right now. Yeah, I think you should get some hint, like, you know, what method you use and if you use case work, like how to divide into... So this is, so yeah, this has been mentioned, this is absolutely a case work problem, but the main thing to realize, and yeah, I'm going to really solving it is that if A1, if, you know, A1 is 1, then the sum of A1 plus A3 plus A5 plus A7 is at least 1, right? We also know that the biggest the sum can be is 4, which is if everything is 1. So if you just have 1s, the sum of 1s is how many there are. I'm going to write that down as if that needs to be written down, but, you know, 1 times a number equals that number. That's what I'm saying. I'm saying that 1 is the identity of multiplication. So the sum of a bunch of 1s is just how many 1s there are. So we're going to do case work on how many 1s. It can be anything from 1 to 4, it can't be 0, because A1 has to be 1, right? So say that there is 1, 1, right? That means that A1 equals 1, and then A3, 5, 7 equals 0. So there's a single 1 there. Similarly, out of A2, A4, and all the way to A8, 1 of them has to be 1, right? So that the sum of this is just 1, and the sum of this is also 1. So if the sum on both sides is 1, there's 1 way to do this here, because it has to be A1. There's 4 ways to do it here, because it can be any of these 4. So that leaves us with 4. There's 4 different ways for both of these sums to equal 1. Then we do the case work on what if the sum's 2? Oh, so I thought for this one, it's only 1, oh, I see. In total, we have only 4, right? Not 8, right? Yeah, in total, we only have 4. So then, if there's 2, well, then A1 has to be 1, right? And then any of these can be the second one. So you've got 3. And similarly here, you just need 2 out of 4, which means we're doing combination, we're doing choose. So we have 4 choose 2, so that would be, you can also think of a permutation, except that there are two 0s, and then there's two 1s that we cancel out. However you want to do it, with the permutation, or with the combination, you still get 4 over 2 times 4 minus 2, right, 4 factorial, 2 factorial, 4 minus 2 factorial, and you get 6, 24 over 4 is 6, so 3 times 6 equals 18. If there are three 1s, well, then A1 has to be a 1, again, it always does. Then two of these have to be 1. So we have to choose which of these two are 1s, which is the same as complementary counting, by the way, choosing which of them is 0. Because if we have two 1s and a 0, that just means we have a 0. The 0 can be either of the three, so there's three ways. Meanwhile if we have three 1s over here, it's the same as choosing one 0, so there's just four ways. That's 12. And that seems kind of wrong, honestly. And then if there is four 1s, then everything is 1, there's one way. That's one. And if you add these together, you get all the cases, right, you have one 1, two 1s, three 1s, four 1s, you get 35, B, which is the correct answer. So, yeah. Yeah, so 35. Yeah, so I think next time we could also maybe do, because I know it's hard for you to draw over here, next time maybe we also try to pull up the, you know, some pre-typed things so that you can explain maybe, just like last time I made a mistake because I was drawing too much on the thing. Yeah, thank you. So, some of you got the correct answer, yes, that's the important thing, like how to divide into case works. And then let's go to the next one, what we have over here. Okay, so Petra has three different dictionaries and two different novels on her shelf. How many ways are there to arrange the books if she wants to keep the dictionaries together and the novels together? So let's see. So we did permutations. Studies are also good questions that also review permutations. Now this one, let's see what we're going to use for this one. We have three dictionaries and two different novels. So we have D1, D2, D3, right, three of them, and we want to keep them together. And we have two novels, N1 and N2, we keep them together. How many different ways we can arrange them in itself? Wow. Yeah, that was an instant. Yes. This is, OK, great. I think this is one of the question, like typical one, if you can get it very quickly. So this one, I mean, the dictionary is that difference. I have a 3 factorial. And this one is 2 factorial. But of course, we can put this one to the left and this one to the right, right? So these two can change the order. So we multiply by 2, actually, 2 factorial. It's just you have the three blocks, and then that the answer is 24, yeah. So usually, when you have a problem, it's likely it could be super quick. But it's here just to make sure that we are comfortable with the formulas. We know what's permutation, what is combinations. So that's great. Looks like everyone's on top of that. But now we're going to go to a question that is a little bit non-standard, but requires you to do something rather than just using the formula. So David, could you please do this question? All right. Peter creates a kangaroo game. The diagram shows the board for the game. At the start, the kangaroo is at the school. Yes, so let me just mark that. Kangaroo starts here, yeah. According to the rules of the game, from any position except home, the kangaroo can jump to either of the two neighboring positions. When the kangaroo lands at home, the game is over. In how many ways can the kangaroo move from S to H in exactly 13 jumps? Notice the ridiculous amount of effort they put into squeezing a kangaroo into this problem. Yeah, so this is the one with, like, constraint. Like, you need 13 jumps. I don't know a quick way to do that. You probably just to play around, you know, maybe. Because I will give a hint. There is a pattern. If you see the pattern, it becomes a lot easier. Exactly, so that requires you, like, just have to, you know, play around with the construct a path and see if you can see something. Try it with, try to get there in two paths, and then three paths, and four paths, maybe, if you're not sure. And then see what happens. See what you notice. Okay. Whoa, we got the right answer very good, very fast. First, if we look at the first jumps real quick, you start at the school, right? Now, obviously at the school, you can either go home or to the playground, but we don't want to go home, right? We have to go home and 13 jumps and we can't go back. So we go to the playground. There's only one thing we can do. It's like one way of jumping. We go to the playground. Once you're at the playground, you can get to either the library or you can go to the school, right? So the first jump, there's one option. You go to the playground, the second jump, there's two options from the playground, you go either to the library or the school, right? Now, if you're at the library or the school, either way, you can only either go home or to the playground. In general, you're either on this diagonal or on this diagonal, alternating. And again, if you're at the library, you don't want to go home. So you have to go to the playground. There's only one way to get there. If you're at the school, you don't want to go home. You want to go to the playground. There's still one way to get there. So the third jump, you've got one option. You go to the playground. The fourth jump, you've got two options. You go to either the library or the school. The fifth jump, you've got one option. You go to the playground and so on and so on and so on. So what you find is you keep doing this until you get to the 11th jump, you go to the playground. The 12th jump, you go to either the library or the school and 13th jump, you go home and you're done. So what this means is that every other jump, you have a choice. You have to make one or two choices. And then for the rest of them, your jump is predetermined, which means that you've got seven odd jumps that don't matter, right? And then six even ones, 2, 4, 6, 8, 10, 12, which means your answer is just 2 times 2 times 2, 6 times, which is 2 to the 6, which is 64. So it looks kind of tricky, but once you find the pattern, that's a very good drawing, by the way, it looks like a DNA helix, I love it. Once you find the pattern that there's 2 and then 2 and then 2, instantly you just multiply by 2, 6 times. And it really doesn't matter if you're at the library or the school, because there is a symmetry line here. Yes. I mean, the question with finding a pattern could be, you know, once you find it out, it's very easy, but, you know, it takes a lot of time and sometimes it's like going to uncharted waters and you don't know where to go. But the method is very general, but it's actually very, very useful. So let's go to the next one. I think, good, we're going to have time to do all of this. So this is a quick question, but I just want to make sure everyone, you know, it's just probably take you 30 seconds, one minute to do it. On the face of the given die, the following numbers appear, you know, minus 3, minus 2, minus 1, 0, 1, 2. If you roll the die twice and multiply the result, what is the probability that the product is negative? I don't think there's a lot of emphasis on this, but sometimes you never know, right, it's just, and it's really just counting because the events are equally likely. Oh, there's still a five pointer, so don't go too fast. Yeah, don't go too fast. Yeah, I mean, So, the gap has to be negative. So, you know, one is negative, one is positive. Rows are die twice, so we have a pair of numbers, right? And the members of the formula, total number of outcomes and number of successful outcomes. How many pairs can you get in total? And how many pairs can you make since the product is a negative number? And we have a poll misclick, and actually that's intended to be the correct one, so. Okay, good. So we know that we have three negative numbers here, right? First, we have, if you want a pair of numbers in A and B, how many different pairs can you make? Now, in this set, it also important if you care about the order or not. In some questions, you can do it either way, either you take it as order of pairs or not order of pairs. Suppose that I care about the order, right? So I cares about the orders and I would have six times six, that is 36 pairs. And then I have a three negatives here. So, and I have a three N and two P, right? So I could have a negative first and a positive later because I care about the order or a positive first and negative later, right? So it just has to be consistent which way you choose. And in either way, I have a three choice for negative numbers, two choice for the positive number. So that's six. And here, I also two times three, which is six. So I have total of 12 successful outcomes divided by the total number of outcomes, which is 36. So the answer is a third, okay? And not a quarter, and not a half. So I said very few questions, but when it comes up, it's really just about counting. Okay, good, let's do another one. We have time for it. The numbers 257 and 338 have the property that when their digits are put in reverse order, they create larger number, 752 and 833. How many three digit numbers have this property? Again, not so much about combination of permutation, but really like how to construct it, really how to count in general. It may take some time to do the calculation, right? Oh, good. Some students are really good. You got it really quickly. Just keep others in the class, you know, a minute or two to get a chance to try it. And David said that you are taking a course at university. And yes, people are learning at university. That's true. You need it for a lot of courses at university. And it's just not covered in a standard curriculum. And that's why I would have a friend who's a professor of physics, and he said to start with statistical mechanics, especially he has to start teaching from the very basic of this thing. And why? The flip side is that if you know this, and you do math competitions, and you're decent, even if you're not amazing at the competition problems, if you know these ideas, and you get put in a high math course that's supposed to be extra super hard, and they're just teaching you math kangaroo theory, it's going to feel really underwhelming. And then you feel really good about yourself. It's funny because a lot of times, because you don't develop the intuition for that, and then we'll go to university and you're bombarded with formula, you just don't have a feel for it. Why? So this is useful not just for competitions. No, it's super useful when you do as I said, statistical physics, computer science. When you need to be notorious, it's super important in computer science. It's just a shame that it's not part of the curriculum. And on the one hand, you say, oh, counting is easy, right? Because the name is accounting. But actually, it's a very difficult subject and super useful. So we can start it a little bit. This is one of the things that if you know how to do, you can do it right away. If you don't know, you just really don't know how to get started. And for me, the easiest way to do is that I would just write out the generalized form for a three-digit numbers, right? We have ABC. There's no constraints. This is ABC. I started with ABC. I mean, B, I can do anything, right? Because when I reverse the order, it just stays the same, becomes CBA. So I really don't have to worry about B. But I have to worry about A and C because the reverse order is number is bigger. That means I have one constraint, C bigger than A, right? And really, with this constraint, I've captured everything about this problem. And now it's easy because I can do casework. It's just the first part is hard. And there's no formula for that. It depends on the problem. You just have to do a lot to build up the experience, right? But once you get to this point, it's easy, right? So case, do casework. When I have A equal to 1, it cannot be equal to 0 because C bigger than A is going to go from 2 to 9, right? When A equal to 2, then C going to go from 3 to 9, right? Does it make sense? And then all the way when A equal to 8, then C is equal to 9. So I just have to count all the places. This should have 8, right? 8 and 7, and plus all the way to 1. So that's 8 times 8 plus 1, which is 9. I divide by 2. That's 36 ways to make A and C. But don't forget about B. B is totally dependent from A and C. And C, B has 10 choices. It can go from 0 to 9, no constraint whatsoever on B. And that's why we have to multiply these numbers, two numbers together. And we had a 260. So I hope just have to do a lot with experience. And you can tackle problems like this. OK, so one minute. So I just quickly, OK, first of all, I want to quickly go through the first, the thing that last time we were doing, the questions. OK, the first one, yes, I just want to apologize. Last time, I was in a hurry because I knew that we were over time. So the last question, when we count, we have to, we asked to find the length of A0, A1. So the last term is actually 2 to the 10, not to the 11. And when I sum up, I was in a hurry. I didn't really listen to what David said. And I summed it incorrectly. So if you did it with the way David did, we have a sign at 83. Or if you are familiar with geometric sequence and you realize that the ratio is minus 2, you can pull up this formula. But really, for math, I think most of the time, you can just calculate it by hand. They don't really focus on formulas. But it's super familiar to know. So we have a few minutes, so I want to introduce one more question, last one, just to get used to many different types. Yeah, so this one. In the 4 by 4 table, some cells must be painted black. The numbers next to and below the table show how many cells in that row or column must be black. In how many ways can this table be painted? Oh, I like this one. Yeah, I know it's really not just, I mean, combination permutation is really important. But there's just so many different other questions. So it's kind of fun and interesting. Clearly, this is a bit of a leap for the other ones. But also pretty clearly, it should probably not be too hard to notice that this is a case work problem. It's just an annoying case work problem. This would probably be a good time to get some paper. Yeah, of course, we probably, it's hard to do this mentally. Oh, no, no, I think you can quickly, I mean, this paper is easy. I'm doing a paper is it just quickly pop it out of the paper, even in the exam, you don't want to write it right on the table, because you after you write once and you know, you cannot do another case, if the number is there, yeah, so it's just quickly. I got the answer is one of the first answer is, let's think about a knee and he could we have more than one to do that right so you know that's that's why I want to pull up question like this so because it could be anything so for this, I would say, the first thing we do that can fill in all the, all the cells that we can write so for example here they all of them are zero. Right. And on the horizontal so here I'm sure I can feel it, you can just fill in whatever cells that you are 100% sure. And then now we have to do case work where else when we feel next right. And I think it's easiest to think about, you know, the cells that has one. So, yeah, okay, let's see, it could be five anyway. So, this one. For example, I consider this vertical one. There's only one so I can probably do four cases, it could be here could be here could be here right. And then I can start with this one, if you feel it up so I have a one here, but it's one then everything else had to be zero. And here is two so this had to be one and one. This one has to be one and one. And this one has to be 00. Yeah, so if the first case I have only one, one choice. And then I can just continue do like that I mean this is just the idea. But the text time is one here and zero is just moves a one up and then try to do in other cases, I think the answer is it. I think probably three. Yeah, I did before but you know it's important thing is idea of how to do how to do case work. You know, if you can have a maybe a faster way to do that but this is the way I do it and I just counted. Just move one cell up. What do you think, David. Yeah, maybe this is a good exercise to try at home. Yeah, but I just want just want to show you that yes, definitely permutation combination and what we discover is super important and you really want to get you with it because it's very useful for level 11 and 12 and when you go to college when you want to continue with math with physics with computer science, but as for math kangaroo, we have all many different types of problems that require a lot of practice and flexibility in your thinking. This is really what math is about, because when you do math at school you learn a bunch of formulas. It's not as much about the thinking but especially with math kangaroo problems. This is where it's up to you in a sense you know you have to figure out the problem yourself. You have to. You're given an entirely new field of problem with every one of these puzzles, and you have to understand it yourself without the prior formula and that's really what problem solving should be about. And I think that's one of the purest parts of this competition.

combinatorics

problem-solving

permutations

combinations

pattern recognition

counting strategies

complementary counting

mathematical principles

real-world applications

mathematics competitions