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Repeat initial order

welcome back to discrete mathematics today we're going to take a look at combinations with repetition so we did combinations without repetition earlier which is just n choose R but in this case when we have n objects and we want to select R of them with replacement it's going to be n plus R minus 1 choose R and I'm going to give you an intuitive explanation why hopefully and with the following example I'll also give you another explanation intuitively so you'll have two different angles to see this so what we're going to do is we're going to take in our example n is equal to 4 and R is equal to 3 so we have four objects here we're just going to call it 1 2 3 & 4 and what we're going to do is we're going to select one of them so for our for our first one we're selecting we're going to take two and we're going to put two into a bin here ok so we selected two so when we select things with combinations we kind of cross it out and say ok we can't take this again but we can take it again so we have to add two back ok now this time where you take let's take 4 next so we put 4 in there we're going to cross it out but because we took 4 out we have to add it back in because 4 gets replaced ok so then let's take 4 again so there we go we select a 2 4 & 4 so there we have it we selected three objects from 4 with repetition so what we really did here is out of five total objects we chose 3 so this is equal to 5 choose 3 now you're saying hold on a second but if I take 1 first and this 2 right here becomes a 1 instead it's a different set not necessarily on the intuition here is that you're just replacing and you're filling in an empty spot for whatever the first one you chose is and then you're filling an empty spot for whatever the second is and of course when we choose the third one we don't have to add a new slot because we're done so that is one intuition of what we're doing here so you can see that this is 5 choose 3 if we put in our so this should be 6 choose 3 if we plug in our formula here we would get 4 plus 3 minus 1 choose 3 which is just equal to 6 choose 3 so that's the intuition here now what can we do well we can say ok suppose we have a donut shop with 20 different kinds of donuts so you know you have chocolate glazed chocolate the Boston cream there's tons of different Donuts um but we want to select 12 different Donuts to take home so really what we're doing here is we're listing out ok we have type 1 we have type 2 we have type 3 all the way up to type 19 and type 20 we do the same thing as before we say ok I want type 2 so we put 2 in our bin we cross it out and then we add an extra slot for 2 so we can take it again so we have n is equal to 20 we have R is equal to 12 because we want 12 different Donuts so we get 20 plus 12 minus 1 choose 12 which is just equal to 31 choose 12 so that's one way of looking at these types of problems another way is this way where we say ok X 1 plus X 2 plus X 3 is equal to 7 all of these are greater or equal to 0 how many positive integer solutions are there to this equation so what we have this time is we have three different bins we're going to call them X 1 X 2 and X 3 and we want to put seven objects into these bins so we could put this one there that one there we're going to fill x2 up with those two and X 3 will get in these three so this will be 2 plus 2 plus 3 is equal to 7 so you're saying okay wait why isn't this just 3 times 3 times 3 times 3 so it would be 3 to the 7 because each circle hasn't has a way of going into seven different containers well as permutations for one and all of these integers they're indistinguishable which means they're all identical so you remember the multinomial theorem where we have like north's and East's and we have to divide by the amount of east factorial and n factorial it's the same idea so instead what we're going to do is introduce this other intuitive notion of ok here's 7 objects 3 4 5 6 7 and I want 3 bins so to make three bins all I need to do is insert two lines anywhere so in this case we have X 1 X 2 and X 3 so X 1 has 2 X 2 has 3 and x 3 has two ok that's not bad um so what if we want one of the bins to have 0 then we just put this line here so this is X 0 and sorry this should be X 1 and this bin X 3 can be here and then X 2 can have all 7 so that's also a possibility so what are we really doing here well really we're taking how many dividers do we need to get three bins we need two dividers so we're actually working with 8 different spots sorry 9 different spots and then we just put our dividers into two of those spots so this is 1 plus 4 plus 2 or we can put our dividers in these two spots and this is just 7 plus 0 plus 0 so you take the amount of dividers you need which is going to be n minus 1 and then you just put them in the extra n minus 1 slots you added so for this equation if we take n plus R minus 1 choose R it's a good idea first to tell us more generally which ones are X and which ones are N and which ones are R so the general solution is that X 1 plus X 2 all the way up to xn is equal to R we'll have n plus R minus 1 choose our solutions where all of the X's are greater or equal to 0 so in this case specifically we just have 3 plus 7 minus 1 choose 7 solutions which is equal to 9 - 7 which is the same thing as saying ok I have 7 objects here but I need two dividers so I'm going to add two more options and I'm going to pick two of them to be dividers so we know that this is equal to 9 choose 2 by the symmetry property so this is the same way as saying ok I just chose 2 of my 9 objects to be dividers so that's the difference here alright let's do something a little bit more challenging that wasn't in the first time I did this video right after this quick example here with a word problem how many ways can we put 10 identical balls into 6 distinct bins well this is the same thing as saying that X 1 plus X 2 all the way up to X 6 is equal to 10 because this right here is the total balls which is like the total numbers and this right here are the bin six distinct bins so this is just 6 plus 10 minus 1 choose 10 which is just equal to 15 choose 10 or 15 choose 5 because we need 5 dividers to get 6 different bins so that was a word problem I rushed through that a little bit but it follows the exact same argument as before so here's something I didn't do in the original video was when one of our conditions has to be greater than another number so X 3 has to be greater than 1 so what this really means is that X 3 has to be greater than or equal to 2 so these are the same thing I'm saying it has to be greater than 1 so it can be greater or equal to 2 okay so let's start up here and say let's let's convert this to a different type of problem let's instead of having objects in a row let's make shelves of objects so we have a shell X 1 a shelf X 2 and a shelf X 3 and I have 9 objects to put on these shelves in fact let's just say they're books so these are going to be our books so the Shelf 3 has to have at least two books so we're going to put two books on the Shelf so what do we have left well we now have seven books left to distribute between these three shelves so really this problem right here is actually just x1 plus x2 plus x3 is equal to 7 where all of our X's are greater than 0 greater or equal to 0 which we know is just the same thing as 9 to 7 from the previous example so here's the question well how do we do that algebraically so let's do this algebraically here I'm just going to copy the question X 1 plus X 2 plus X 3 is equal to 9 where X 3 is greater than or equal to 2 okay what we're going to do is make a substitution we're going to say X 3 prime is equal to X 3 minus 2 so that way X 3 prime is going to be greater or equal to 0 so what we get is we get X 1 plus X 2 plus well hold on we have to substitute X 3 n so X 3 is the same thing as X 3 prime plus 2 so X 3 prime plus 2 is equal to 9 so we just substitute it in X 3 prime plus 2 equals 9 because we want to shift this to here to a 0 so we made the substitution possible and now we just use some algebra here so X 1 plus X 2 plus X 3 prime is going to equal to 7 because we just put the 2 to the other side and then we just find the solutions for this one because all of our X is are now greater than zero which we know is just 9 2 7 or 9 choose 2 if you look at it from a divider way so that really is just combinations with repetition it seems challenging at first and it can be challenging because there's different ways of looking at it there's balls and bins you can always do that way or you can do books on shelves you can do the integer solutions and when you get to discrete math 2 and cover inclusion exclusion you can put upper bounds on those into integer solutions too so we can say that X 3 has to be less than 9 but that will be covered now the next video will be a video with tons of practice problems on combinations with repetitions just regular combinations and permutations so if you have any questions you can leave them in the comments below or you can check out TREB too calm and there's more stuff there we also have a subreddit reddit.com slash are slash trip tutor and you can also post questions there and I will be there responding so as always have a great day and I hope to see you guys next time