Transcription
hello everyone. In this class, we'll be going over many topics, including accounting and probability, number theory, algebra, and geometry. These are many topics that are really useful for the AMCA and become a very oxen.
Today, we'll be going over counting, probability, specifically permutations and combinations. Let's start out with today. Can you all see my screen okay, and you can hear me right—right? Okay, so then we'll get started today. So let's start the first problem.
If you examine one—if you have Alice, Betty, and Chase, and you need to choose two to be the president and vice president of the math club, how many ways are there to do this? So how do we approach this problem? One way is just to list all of them up. So as you can see here, I've listed all them out. So we have to do based on the different people who can become president and vice president. So as Alice is the president, then we can have either Betty be the vice president or Chase be the vice president. On the other hand, if Betty is the president, then either Alex or Chase can be the vice finished, but then it's changed the president, which is not either Alice or Betty be the best. So as you can see here, there are six total ways that this will work. But how can we do this without listing all the?
Let's consider all three people to be A, B, and C. So for the president, we have three choices; it can be any one of them; it can be any one of A, B, or C, because the president could be anybody. So there are three choices for who could be the president. Then for the vice president, it can be any any of the people who are not already chosen to the president. So let's say the president is chosen as A; then you can cross out A. This Betty no longer be the vice president as well. Therefore, there are two choices for the vice president because it can be any one of A or C. Now that they're the ones were not chosen that to test it. Therefore, there are six ways to choose president and vice president from three people, and as you can see this is the same answer that we got. Let me just list it all the possibilities out. So do any of you have any questions about this problem? You can put them in the track to have any questions. You okay? There are who wants next problem?
So just like the first example, this is a very similar problem. Find the numbers ways of selecting a vice president, president, and secretary from the group of eight people. So what is it? Let's first think about—let's let the people be one, two, three, four, five, six, seven, and eight. So these are the eight people that we have. And just like the last example, president can be any one of these people, so there are eight choices for who can be the president of the club. And now looks safe to say that six gives the president that was chosen. Now you only have seven people left, so we have only seven choices for who we can choose as the vice president. Now let's say we cross out one more person; let's say the vice president was—let's say the vice president was three, for example. Now there are only six people left: one, two, four, five, seven, and eight. So then we can only have six choices for who is the secretary of the slub. Therefore, there are only eight times seven times six ways to choose the pregnant, vice president, as secretaries from the group of eight people. This is just 336 if you multiply it out. So this is the final answer for this problem. Do any of you have any questions on this problem? You can put them in the try a few. Do you—you? Okay, now I'll move on to the third example.
So find a number of three-digit numbers, all of us students distinct. So three-digit number has three different digits. Let's draw out each of the three digits that we had. So of course the three even each of the digits can be anything from zero to ninety except the first two jail. So the first digit can be one, two, three, four, five, six, seven, eight, or nine. Note that the first digit cannot be zero since if the first of your three-digit number was zero, it would no longer be a three-digit number; it would be a two-digit number. So the digits of this place you know may be 1, 2, 3, 4, 5, 6, 7, 8, & 9. So there's only have nine choices for this digit. Let's just say this digit flows for example in size. Now nobody wanted Mexican. And notice here that the problem says finding on those three-digit numbers that have all my students are distinct, and basically what distinct means is they have all different. So for example, one two three would be a valid number, but one one two would it because these two digits right here, as you can see here, these two digits are the same. So that's not allowed because we've not all the digits to be distinct or different. So I forget to here if this is a five, this can be anything butterflies. So this can be 0, 1, 2, 3, or 4, but then it can be a 5–OH only the first two days is 5, so it can be 6, 7, 8, or 9. So there are 1, 2, 3, 4, 5, 6, 7, 8, 9 is for this this place now. So the final did it. Now we've already got to say this digit right here was a 4, for example. Now these to do their difference; that's good so far. And I was remaining digit again can be anything from zero to nine shooting 5 & 4. 0, 1, 2, 3, 4; we can't have a force and look not 4 or 5; we can have a 6, 7, 8, or 9. So there are—so there are 8 choices in this case and this next a 8 2 3. So then a 543 would be a valid number. So just to summarize, the first eject cannot be 0 of course, so then it could be anything from one to nine, and there's nine choices with that. Second digit can be anything from 0 to 9, for example of course it can be 0; 106 would be a balance over because the second digit can be 0, but it can be the number that was in the first digit. So then we subtract 1 from 10 because there's tendance to 0; it's not. So they're only going to be 9 sources for the second digit. And for the final digit, we can't have the same number as any of these two digits, so therefore it has to be any number from 0 to 9 besides the lid that we chose the previous two digits. There are eight choices for this, and if you multiply this out you get that there are 648 ways the 3em3 two numbers with all of it suggests this date. Okay, so do all do it if you have any questions you can either unmute yourself or you can put them in the chat. You—you—you. Make sure you understand how to solve this problem because the problem the palms we're going to do from now on carding involve the concepts that we learned here. Okay, I will move on then.
So let's introduce factorials. Basically, a factorial is a number multiplied by all the numbers less. So n factorial would be n times n minus 1 times n minus 2 so on all the way sometimes one. And even though this may seem confusing at first, it really is very simple. For example, 4 factorial is just 4 times 3 times 2 times 1, which is 24. And for another example is 3 factorial, which you just be 3 times 2 times 1 or equal to 6. So in the chat—in the chat, put what you think the value of 5 factorial is. You yeah, there is 120. I factorial is just 5 times 4 times 3 times 2 times 1, which is 120. So here's some other time factorials it might wanna memorize. 0 factorial is 1, and I know this might seem strange at first, but it's an exception to the factorial, so you might just wanna memorize this. And again, 1 factorial of obviously 1, 2 factorial will be 2 times 1 factorial just 2, 3 factorial will be 3 times the value of 2 factorial; I'm just going to be 6 + 4 factorial as you type it in here; this is going to be 4 times the value of 3 factorial, just 24. And 5 factorial is 120, and 6 left totally will just be 6 times 2 value is 5 factorial, so it will be 720. So now try an example. Find the number of ways to arrange 5 different books on a bookshelf.
So let's start with the each of the values. So for the—let's just say we have 5 slots: 1, 2, 3, 4, 5. And let's say our five books are A, B, C, D, and E. So these are our five books. So how many choices do we have the first spot? Well, it can get any one of A through E, so there's gonna be five choices for this. Now I'll just say that A was the first book that we kept here. Now we have four choices left, so we have four choices for the book that can be the second slot. So there's it's going to be four choices for this because now you cannot have same—you cannot have the same book in first and second floors. And now for the third slot, you can't have any of the other proceeding book. So let's just say that C was the one that means for the second slot. So this would be A, this was C. So now for the third slot we can have A or C, so we only have B, D, or E, or maybe they're three twenties for this. And now for the final—for the final two slots—for the—for this fourth slot here, let's just say that this value over here was B; 70 cross out B if you can't use anymore. So we only have two choices for this second book here. So there's two choices for that. To not have to say this value is E; let's to say. So we can cross that out. And now for this final slot here—for this final slot here, we only have one choice; this is only one looks left, and this will of course be E. So therefore, it's just the number of ways to range five different books on the bookshelf is just 5 times 4 times 3 times 2 times 1. And if you notice this is exactly the value of five factorial, so therefore this is just going to be equal to 5 factorial, which if you memorize to sum up it's just going to be 120. Okay, so do you all understand how we got 120 for this? Could you look at the number of ways—number to book choices for each of the five spots? Okay, now let's move on to the next problem, which is a word or a trick problem. So I'm going to give you a minute or two to try and solve this problem on your own. Don't put it in the comments at please, and other people can see your answers, just a directly private message it to me. So how many ways are there to range letters of the word not? So I'll give you around ten or so seconds to try and solve this. You okay? Yes, the answer is 24. It looks like awesome, you got that. So this is very simple; there are just four letters: N, O, T, and H. So let's say there are four slots in which we have to put the four letters in because that's how many were in two letters; you're just going to be four choices for the first book, three choices for the second book cuz after we cross out one, then two choices with the third book, and one choice for the final book. This is just four factorial, and it's 24. A lot of you, but now let's try a slightly harder example. How many ways are there to range the letters of the word Mississippi?
So Mississippi has one, two, three, four, five, six, seven, eight, nine, ten, eleven letters. So we have 11 factorial ways to arrange them, but then you might notice that these—you have multiple repeat letters. So essentially at this four I's as you can see. So straight them out in fashion or you can see the duplicates. So we have M, S, S, S, S, I, P, P, I, I, I, I. So as you can see there are four I's fluorescent and two P's. So does anybody know of what we can do to account after four I's? Note that we can just—if we were to do the four I's, we can just rearrange the four I's. So for example, in arrangement if you had N, I, A, M, I, I, and then we had S, S, I—P, S, S, let's say, then having this I here and this I here it's the same as having this I here and this I here, so we are over topping for this. Is anybody know how we can account for this? We can divide it by four for all of the I's, not but because there are four factorial ways to arrange I's, not just four. So we actually divide by 4 factorial ways because think about that the I and see there, I1, I2, I3, and I4. So we can just essentially rearrange all of these in four factorial ways. So we divided by 4 factorial, as you can see here, and similarly you don't do for the S's. If I xi f multiply 4 factorial in the denominator by 4 factorial, yeah. So we multiply by another 4 factorial here and similarly for the P's. Yeah, sure, just feet, you can speak or you can type in chat, divided by 2 factorial. Yeah, you can just divide by 2 factorial. And now all you have to do is evaluate this, and I'm not going to do all the calculations because this is going to be very tedious, so we can just leave our answer like this for now. So now let's move on to the general formula for the number of ways to arrange the letters of a word. So for basically the formula is a factor you'll over D1 factorial, D2 times E2 factorial times Et sectoral so on. And basically, n is the number of objects or age or words, letters in the word, and under in the caboose example, and do you want, do you to teach three or number of clients each of the letters that occurred more than one time here the word and it was amazing a bit complicated at first, but we can just try it but just use the formal on this example, pineapple.
So pineapple, yeah, we list letters of the word pineapple. So we see that there are a total of one, two, three, four, five, six, seven, eight, nine letters of the word. So by the formula here, n is equal to nine, so we just plug in at that. Now we see the D1, D2, D3, those are number times—the letters that are computed one time periods work. So you must divide by that to account for the rearrangements of this word. So as you can see there are three P's, two E's. So from our formula, we have this will be nine factorial over 3 factorial because there's three P's and two factorial because there are two E's. And I'm again, this can simplify this because it's just calculations that you can appreciate you to do on your own. So the answer would just be bits whatever it is always—. And again, you can just evaluate nine factorial to be nine times eight have seven and six times so on, and do check tutorial is you memorize the earlier of six and two secondly would this be— —just really just evaluate this. So now let's move on to permutations. Permutations is a number of ways of rearranging things when the order matter, exploring is important. So let's try this example. How many ways are there to particular favorite and second favorite book books? I saw on the previously the very beginning examples. Let's just use a method where he has two books, the favorite and second segment. This would be the favorite book and says to be the second favorite book. So because there are any it can be any one of five books, there are going to be five closest to the favorite book. Then we have to cross out one because you can't even—you can't have the same book even favorites and a second season. So then they're going to be four choices from second later. Now let's forget to the formula for signing K distinct positions to ed thanks. So basically the formula that we can apply to these types of problems. So this one is P and K, and don't worry about the sensation; it basically just means the number of ways of permuting—number ways to—it just basically means the number of ways to sign can see positions to end things. So if you want to ignore this for now, but if it's just a notation that's used, so don't worry about this too much; just remember that this is the formula, and it's n factorial over N minus K factorial. So let's explain this with the same example here. So instead of doing this method, we can just—we can try using the front left to solve the same problem. So in this case, you have since there are a total of five books yet that n is equal to 5, and how many distinct positions do we have? Well, two positions that we have or the favorite that means because K is equal to 2, that makes P of n comma K is going to be equal to 5 factorial over 5 minus 2 factorial, which is equal to 5 factorial over 3 factorial, which is equal to 5 times 4 times 3 factorial. Notice that I just put the 5 factorial into 5 times less each factorial, so it will cancel out with the denominator and not on the waiting it because it will just cancel out. And for 3 factorial, these terms cancel and we get an answer of 20, just like we got with this method. So look, let this work except this is just the forelock for general you solve it all these kinds of problems. So here's another problem. Again, this notation just means basically if you have six objects, you need to choose three positions, and similarly for 7 & 2. So I'll do the first example for you, and you guys can try the second. So the first example we just—this is a formula for your best to refer to. And so P of n of K, so n is equal to 6 here and K is equal to 3, so this is going to be equal to 6 factorial over N minus K, and n is 6, K is 3; 6 minus 3 factorial, which is equal to 6 factorial over 3 factorial, which is equal to 720 or 6 or 120. And instead of doing this, we could have just also done 6—pack 6 times 5 times 4 & 3 factorial over 3 factorial to just also me 120. Well, and now here's another problem: evaluate p7 comma 2. So I'm going to give about 30 or so seconds to try this problem on your own, and if you got conceives you can do it. You—you—[Music]—you—you—you. Okay, so here we know that n is equal to 7 and K is equal to 2, so now I'll just try putting it into this formula over here. Okay, so I'm gonna write a couple more steps. So he has seven looks like both looks like a five or six people got it so far, so I'm going to give one more hint for this problem. So we have 7 factorial over 7 minus 2 factorial, which is equal to 7 factorial over 5 factorial. Now this is very simple; just equal to 7 times 6 times 5 factorial over 5 factorial, and these terms cancels; we get an answer 42, which is what men up to you God. So do you all understand how to do this? Okay, now let's try a very similar problem and also apply the formula that we use to a sexual problem. There are ten students at the math contest. How many ways in there to sweat the first, second, and third place winners? You—you. So they anyone want to explain how they solve this problem? Yeah, sure. Okay, so first I know that there are 10 total students, so first I wrote 10 factorial, okay, and then look—so down numerator of the fraction now I—now that I know there are three places, so I would have to do 10 minus 3 that's in parentheses, and I would have to find the factorial of that. I can simplify this to 10 factorial over 7 factorial, which and 10 factorial can also be written as 10 times 9 times 8 times 7 factorial, and the 7 factorial in the numerator and denominator both cancel out, so then I do 10 times 9 times 8, which is 720. Yeah, essentially this is just 10 P 3; this is because we're treating three positions for the 10 students. And now she said this is just going to be 10 factorial over a 10 minus 3 factorial, which is...
Going to be 10 factorial over 7 factorial, which is 10 times 9 times 8 times 7 factorial over 7 factorial. These cancel; teachers cancel to get an answer of 720, which is correct. So now let's move on to combinations.
Example 11: If you have Alice, Many, Chase, and David, and you need to choose two of them to be the leaders of the math club, how many ways are there to do this? So we can just list them out again. So you can either have Alice as leader one and Many as leader two, or if the leader one is Alice, I can have Chase, or if it can have Many, Chase. You need Alice as leader two; if the leader one is David, you can have Alice, Chase, or David has Many; and if he has Chase, this will either one. You can have Alice be a leader; and if you have David as leader one, another will study it will Chase or sweeter.
So do we notice that anything wrong here? You see that Alice and Betty, or Betty and Alice, these two cases they're the same thing because there's no such thing as either one or two; just up there, both leaders. These two phases are the same thing. So if we flip Alice and Betty, or Betty and Alice, you get the same exact configuration. So sweeping these two, these two people's places doesn't really make a difference, and the same same thing for all the other cases for Alice and Chase over here. This case over here will also be the same; it's Chase and Alice also over here. Similarly, this case over here, Chase and Betty, this may be the same as Betty and Chase; you're just being leader one or leader two; the same thing. And similarly for all the cases; similarly for all the cases, we we can just switch them up and they'll be the same thing. So therefore, we have to divide by two from our trouble count.
So as you can see, we have enumerated 12 ways here, except that each way is over-counted because each way is to conduct one person being leader one and the other person being leader two, or having one person be leader one and the other person's view later lot there's one. We have to divide by two; three minutes we have 12 divided by two, which is equal to 6. Now how do I solve this problem without listing all the cases like this again? We afford to insist for two less proved to be the leader one and three choices for me to be leader two, but then we have to divide by two, being a leader one same thing as being leader two; we can just swap them, drop their places, and these only get an answer of 6. But the correct way of thinking about this should actually be two factorial. This is the class there are two factorial ways to rearrange the people. Therefore, for example, if you have to choose three leaders instead of just two leaders, instead of the just three would be three factorial because there's three factorial ways to arrange those three people in the lion's head. For example, I'll choose three leaders; you can have ABC, for example, or we could have ACB, or we could have BAC, or BCA, or CAB, or CBA. All these combinations are the same, and there are three factorial ways to arrange these three people in the same order, so we would have to divide by three factorial, so they're just two factorial.
Now let's move on to the general formula for choosing objects. Since formulas are choosing K objects from n objects, just n choose K. Also, this notation here is just like the P notation, and this invitation is very common, so you just should just memorize how to like this. This time is write it, you know, in a parenthesis; you just write n on top and k at the bottom, and that basically just means n choose k. So basically, the formula is n factorial over k factorial times n minus k factorial. So and also let's evaluate 8 choose 4, just to give you an example of how to do this. So 8 choose 4 would be 8 factorial over 4 factorial times 8 minus 4 factorial, just from this lemma because we have n is equal to 8 right here and k is equal to 4. So we just plug in k here, plug in n here and here, and then we just get that x equal to this expression; just going to be equal to 8 factorial over 4 factorial times 4 factorial, and we can rewrite this as 8 times 7 times 6 times 5 over 4 factorial times 4 factorial here, and 4 factorial and these need just cancel out. This is going to be equal to 8 times 7 times 6 times 5 over 24. These two terms, 48, we can catch these out, two here, and this product is a product of 70; would be the answer to 8 choose 4. So 8 choose 4 is equal to 70. And you also might notice that n choose k is equal to n choose n minus k; xi is equal to n choose n minus k. Sum of all this is because that if you have k factorial, just to give an example, all you six n is equal to 6 and k is equal to 2, so we have 6 choose 2, and this is going to be equal to 6 factorial over 2 factorial times 6 minus 4 factorial, and then we have 6 choose 4, which is going to be 6 factorial over 4 factorial times 6 minus 4 factorial, and this is also equal to 6 factorial. As you can see, that this is going to be 6 factorial; it should be 2, 6 factorial over 4 factorial times 2 factorial. This is going to be 6 factorial over 2 factorial times 4 factorial; these are just going to be equal. The combinatorial explanation for this is that the reason is this because we can just choose k things to be part to be chosen, or we can also choose n minus k things not to be chosen. So for example, if you have 6 balls, we have to choose two of them would be the same as choosing 4 of them not to easily. So therefore, this is the same thing. And in general, distinguishing between combinations and permutations can be tricky. So the words permute, order does not matter; in other words, apply a permutation, but in the words choose, select, order does not matter; these typically means combinations. So I've already gone over 8 choose 4; try to figure out the value of 9 choose 7. Under you um can you move up so that yeah, thanks. You can see the formula. Okay, thank you. You you you you you you you here, living in the formula, we have n is equal to 9, k is equal to 7. It's not try plugging it into the formula. You okay, now we're going to move on. So we have that 9 choose 7 is equal to 9 factorial over 7 factorial times 9 minus 7 factorial, which is equal to 9 factorial over 7 factorial times 2 factorial, which is equal to 9 times 8 times 7 factorial over 7 factorial times 2 factorial, and the Senate rules will just cancel out. We have this is equal to 9 times 8 divided by 2 factorial, which is 36. Okay, now we're going to move on to another example that involves choosing.
So how many ways are there to select three foods amongst eight different fruits? I'll give you about 30 or 40, thirtyish seconds to work on this problem. And again, this is just be a just an application of choosing; try to use a formula for choosing on this problem. Again, you you it's the other hit; could be the choose three different fruits from eight fruits. All you do is evaluate the value of 8 choose 3. You you you you okay, I'm gonna move on now. So in this form, we have n is equal to 8 and k is equal to 3. This is equal to 8 factorial over, by weakside, 3 factorial, 3 factorial times 8 minus 3 factorial, which is equal to 8 factorial over 3 factorial times 5 factorial, which is 8, 7, 6, 5 factorial over 3 factorial times 5 factorial, and these factorials cancel out, so you have 8 times 7 times 6 divided by 6, which is equal to 56. So now let's try some some slightly higher problems. How many three-digit numbers are there with all the digits odd? So basically, we can have any number for each of the three digits; can be one, three, five, seven, or nine because all the digits have to be odd, so they're going to be five choices for this block, and the same thing for the this ball can be one, three, five, seven, or nine, so it's going to be five choices, and similarly here, I have choices. So this is a product of 125; to their final answer. And now let's try a very similar problem. How many three-digit numbers are there with all of its digits even? So try to solve this problem; honey, I want to see if you can do it. So again, just try three things and make sure to be careful; all the digits has to be even.
Okay, so I think a lot few caught it now. So here's it's very classic; an urgent problem. You just think there's five choices for the first digit, 5 to the second and 5 to the third, just like you need a box except there's one thing that you're missing here. To do this, the first thing it can be 0, 2, 4, 6, or 8, but it can be 0 because if it's 0, then this two-digit number. So instead, it actually is only four things that can work; 0 does not work because 0 cannot be the start of a two-digit number, so there are four choices for this. So you can multiply this out, and again, that 4 times 5 times 5 is equal to 100, which is the answer here. Make sure to remember about a 0 cannot be the beginning of a number, and yeah, in fact, it's not the same answer is the odd case. It's now we're gonna move on to this problem. This is a slightly a slightly more tricky problem. So how many three-digit numbers are there with all the digits distinct? So again, let's try it's a method that we've been using for the previous problems. This first number can be anything from 1 to 9, so there are 9 choices for this, and on the second number cannot be the same; it can be anything from 0 to 19 except the one that's already here. So if you choose 6, for example, can't be 6, so there are only nine choices for this, and now we see that's a third digit. So the surgeon that has to be odd; we can't use any of the numbers already used, but we see but we don't know how many odd numbers real reviews this could be 6, for example; this could be 4, or this could be 5 and 3, and each of these different cases you could get different number of ships that actually work with the third guy, so this might be a little bit more tricky, and we clean up these caterer until this goes next week, but instead. So there's a much simpler way that you can go about this. Instead of starting from the beginning, let's try starting at the end instead. So for this last digit, it can be anything 1, 3, 5, 7, or 9, so there are going to be 5 choices. Realistic enough to save you choose 3 fruits, so we know 3 is used, and now we can go back to the middle digit, or we can go back to our first date, which is probably better. So for this first date, we can have either 1, 2, 3, 4, 5, 6, 7, 8, or 9, and since 3 is overview, you can have 3, so you have 8 choices for this digit, and now for this final digit, there are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; we can have you slip to see me select on board for this case 3; you can ou have 4 or 3, so there are going to be 8 choices for this. The statue has a product 320. So in general, for most counting commas in which tend to multiply for each of the values, it's generally best still deal with the most restrictive conditions first. So for example, this condition would be the most restrictive because you can only have 1, 3, 5, 7, or 9, but then this digit would be the second most restrictive because we cannot have 0, and this is it right here. This would be the least restrictive because we cannot have because we can have any digit in the 0 to 9, so it's always best deal with the most restrictive cases first and the least restrictive cases last; does that makes it easier for us, and as you saw, would be a lot harder if the actually started with the digit first because then we run into some issues; I'm having different cases, so you have if you have any uh you have any questions about this problem. Ok, so now I'm going to kids next problem. So just like the word problems that we've been dealing with earlier, except it's slightly different. How many ways are there to misspell the word misspelled? So I'm just going to write it out in a pattern where you can see the duplicates; miss. So here we can see that there are 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 letters in the word, so we have 10 factorial on top, and now on the bottom, how do we put you try to finish this problem on your own. You don't it's okay if you don't simplify your answer for now if you can sleep in terms of factorials or whatever you have. You you you you you you. So do does anybody want to know you want to say how we can proceed? So for but I've been looking at the bladder, so what do we go back to this? Just put in the trap on a pristine; you can bring the common cat for now because yeah, so it is two factorial for the Xs. Now does anybody want to tell me what we can divide cities? Yes, we can divide by two factorial again, and same thing for the Ls and its forces, and many people actually miss this in the first try, but they just thought this was the answer right here for them to repeat the product closely. They asked for the number of ways to misspell the word misspelled, so this actual original permutation itself does not work. We actually just subtracted one; this is because if you have the same exact thing, you're not misspelling it; it's still correct. So actually, the answer would be this and not just but not just this part right here, so it's gonna be this whole thing combined. Make sure not to miss this as this is one of them, one of the main mistakes that is very common on this check these types of problems because if you have the same letters again, and that's not considered as misspelled; it's still a correctly spelled word, and in order to misspell the word in this word, it needs to be different permutations from this one right here. Okay, so I'm going to move on to prom to the 2018 AMC 8. Professor Chen has 9 different language books lined up on a bookshelf; to array big three German and four Spanish. How many ways are there to arrange the nine books on the shelf, keeping the Arabian books together and keeping these Spanish books together? So you have nine bucks; that's the two Arabian books be a1 and a2, and again, remember the two Arabian books that doesn't mean they're the same book, so we're left to be a1 and a2; the three German books be G1, G2, and G3; in the four Spanish books, well, they can be s1, s2, s3. So it says how many ways are there to raise nine books keeping these Arabian things together and the Spanish works together? So based on our conditions, you have to keep these books together and these books together, so lets them just be one whole mass like this, since they anyways have to be together; just let's just make this call call this big A; let's call this lowercase a and has all the read big books in it; let's call this just I know you think so much to do not but cool group of I'm just reading symbols to know if the whole group. Well, I'm sure this there's actually four Spanish books; this is gonna be four Spanish books, so there's gonna be all these books are donated by lowercase s; gonna be the whole group of books, and these German books you can just keep them in the original form since there's no condition from the problem saying you have to keep them together, so they're just G1, G2, in G3. So the number of ways to reign to all sides of these objects is just going to be 5 factorial because there's 5 objects and need to raise the decline, and then we must divide; they must also multiply the number of ways to arrange each of these books inside each of these poems to me each of these groups. You called an A and s3 to find the number of ways of arranging books inside these groups to learn new ways of arranging them inside this group would just be 2 factorial, and so the number of ways of arranging them inside this group would you just be 4 factorial, so you can multiply this out, and we get that the answer would be 120 times 2 times 24; you give an answer of 5,760. And this was a little bit of a tricky problem because we had to take care of two cases: number one, we have to group them all together; we have to group all these books together right here, and also to Gupta; we also to cook theoretic books together here, so and then you have to order in terms of things inside each of these groups, and then if you multiply them all together, and we get this answer of 5,760. Now let's try now see a little bit of extra time; I'll go over another side; this will work; number five doesn't work puzzles. So a special type of license plate includes three distinct letters in the beginning and four single-digit numbers. How many such license plates exist? So try and go ahead and try this problem, and it's very similar to the comments that we've already been doing, and if you want to kiss a calculator or just leave your answer in simplified form for this problem here. So I'll just go ahead and draw the beginning here: three distinct letters and four numbers. Blue is a letter; two the numbers, and of course, it you know there's 26 letters in the alphabet, and yeah, so try to try doing this problem. Yeah, we're gay; just just to be clear, this is the problem that we're doing right here, number five. You you you you you you you I mean it. So for this first bring his first letter here, they're going to be 26 choices, and for the second letter here, they're going to be 25 choices because we've already used one letter, and now for this final choice here, they're going to be 24 choices because you cannot have any of the two previous letters; we cannot have k or l; she's going to be 24 times goodbye and single unique numbers; that's very straightforward. So single easy numbers, they're just going to be ten choices; treats of them, and that's just because there can be any number from zero to nine; those are just ten numbers, and same thing for all four of these choices. When we multiply it out, you multiply 26 times 25 and it's 24 times 10 to the power of 4; we get that the answer is of 1,561,560,000. And followed that's followed by six zeros, and yeah, you can use a calculator for this because do you want to or yeah, you don't have to; you actually evaluate this have you don't want to, and you get the answer this 156 followed by six zeros; that's the answer for this problem, and if you want to just left it at this form because it's okay for now. Okay, everyone, thank you for coming to this class, and I will send you the the 500 contact you have for this week and try to do the best and also send a submission form where you can submit your answers to all these problems. Okay, bye.