Hey, everyone. We now know how to find permutations when dealing with multiple different or distinct objects, like picking 5 different shirts to wear over 5 days or choosing 6 marbles out of a bag of 6 different colored marbles. But what if instead in that bag there were 4 blue marbles and 2 red marbles? If I chose this blue marble on the first try, wouldn't it be the same thing as if I were to choose this blue marble? Well, yes, it would be the same thing, which means that we're going to have a different number of outcomes here. So when faced with multiple objects that are the same or non-distinct, we're going to have to go about it a little bit differently. Now just saying that permutations of non-distinct objects sound really complicated but you don't have to worry because here we're just going to take our equation for permutations that we already know and just alter it slightly in order to account for these objects that are the same. And here I'm going to walk you through exactly what that change is and how we're going to use this equation. So let's go ahead and get started.

So when finding all of the outcomes of drawing 6 different marbles out of a bag of 6 different colored marbles, simply using our permutations equation, we're going to take our total and the factorial of that and divide it by our total minus the number of things that we're choosing and the factorial of that. Now since the number of things we're choosing turns out to be the same thing, I get 6 factorial over 6 minus 6 factorial using this formula here. And when looking at our permutations of non-distinct objects with our bag of 4 blue marbles and 2 red marbles, we're going to start out the same way. So still taking our total number of objects and the factorial of that on the top. Now here, since I still have a total of 6, I get 6 factorial. Now in this case, we have blue marbles and we have red marbles, 2 different types of objects. So we want to look at the number of different types of objects that we have and take the factorial of that on the bottom. So since I have 2 different types of objects, I'm going to make sure and count those and take the factorial of the number of each of them on the bottom. So here since I have 4 blue marbles, I'm going to take 4 factorial, and then I have 2 red marbles so I'm going to multiply that 4 factorial by 2 factorial and this is my final equation here and I would just solve that as we already know how. Now that we've seen that equation, let's go ahead and work through some examples to make this a bit more clear.

So looking at our first example here, we are asked how many different 8-digit codes can be made from 5 zeros and 3 ones. Now the first thing we want to do here as we did with permutations for distinct objects is still identify what our value of n and what our different values of r are. So here remember n is your highest number it's your total. So in this case I have a total of 8 digits so this represents my value for n. Then since I have 2 distinct objects, I have zeros and I have ones, that means I have this value for r1, which is 5, and then I have this value for r2, which is 3. Now it doesn't really matter which one is r1 and r2 so long as you're accounting for them both. So let's go ahead and write our permutations formula following this here. So we're going to again take our total on the top, our value for n, so 8 factorial, and then we're going to divide by 5 factorial times 3 factorial. Now, here, we've been rewriting our numerator in order to cancel our denominator. Now since we have multiple factorials on the bottom, we are simply going to rewrite our numerator to cancel the highest factorial in order to simplify this in the most effective way. So from here, we're going to want to cancel this 5 factorial, so let's rewrite 8 factorial, our numerator, in order to cancel that. So I can rewrite 8 factorial as 8 times 7 times 6 times 5 factorial. And then in my denominator, I have 5 factorial and 3 factorial. Now those 5 factorials are going to cancel leaving me on the top with 8 times 7 times 6 and in my denominator with 3 factorial, which expanded out, is simply 3 times 2 times 1. Now we could choose to cancel some more things here because you might see a couple of things that cancel but I'm going to go ahead and just fully multiply this out in order to simplify from there. So on the numerator I end up getting 336 and then in the denominator I get 6. Now dividing this out I end up with a final answer of 56 different 8-digit codes that contain 5 zeros and 3 ones. Let's take a look at our second example here.

We're asked how many different ways there are to arrange the letters of the word "banana". Now this might seem like a kind of strange question, but we can still attack it the same way. Now the first thing again we want to do is identify our different values here. So we want to find n and we want to find any values of r, r1, r2, and if I have more than that as well. So here, since we're looking at this word, "banana", we want to know how to arrange all of these different letters. So that means my value of n is going to be the total number of letters in this word. Now the word "banana" has 6 different letters, so that tells me that my value of n is equal to 6. Now we need to identify all of the different types of objects that we have here, in this case, any letters that are the same in this word. Now starting out with b, that's going to be my first type. There's only 1 b in the word "banana", so that means my value for r1 is simply going to be 1. Then I get to my a, and there are actually 3a's in the word "banana", so that tells me that my r2 is going to be 3. Now finally, I have n here. And there are 2 n's in the word "banana". So r3 is going to be 2. Now from here, since we have our n and all of our values of r, we can go ahead and use our formula. So I'm going to take that n factorial on the top as I always do, so 6 factorial. And then I'm going to divide that by all of my different types of objects and the factorials of those. So I have 1 factorial times 3 factorial times 2 factorial. And remember, we're going to rewrite the numerator in order to cancel the highest factorial in the denominator here. Now, my highest factorial here is this 3 factorial. So we're going to go ahead and rewrite that numerator to cancel that out. Now doing that, I get 6 times 5 times 4 times 3 factorial. And then in my denominator, I still have that 1 factorial times 3 factorial times 2 factorial. Now those 3 factorials will, of course, cancel, leaving me with my numerator as 6 times 5 times 4. And then in my denominator, I know that 1 factorial is just 1, so it's not going to do anything there. And then 2 factorial is 2 times 1. Now fully multiplying that out, that gives me 120 in my numerator divided by 2 in my denominator to give me a final answer of 60 different ways to arrange the letters in the word "banana". Now that we've seen how to work with permutations of non-distinct objects, let's get some more practice. Thanks for watching, and I'll see you in the next one.