Combinatorics - Online Tutor, Practice Problems & Exam Prep

The fundamental counting principle allows us to determine the total number of outcomes when faced with multiple choices. For example, if you have 3 shirts and 4 pairs of pants, the total outfits can be calculated as $\left(3\right)\left(4\right)$ = 12. When considering arrangements, permutations account for order, while combinations focus on selection without regard to order. The formulas for these concepts are essential for solving problems in probability and statistics, enhancing understanding of outcomes and choices.

Fundamental Counting Principle

If you have one clean shirt and one clean pair of pants, then you have a super easy decision when getting ready in the morning. But what if you just did laundry and you actually have 3 clean shirts and 4 clean pairs of pants? How many different possible outfits could you come up with then? Well, this is actually the exact sort of thing that you'll be asked to do: find the number of total possible outcomes when faced with multiple options for multiple different things. This might sound like it could be overwhelming or time-consuming having to come up with all of these, so let's go ahead and just jump right in here. Now with our 3 shirts and 4 pairs of pants, if I take that first color shirt and pair it with all four different colors of my pants and then do the same thing for my 2nd and 3rd shirts, I see that I have 12 total possible outfits that I could make on that day. But I can actually come to this conclusion much easier using the fundamental counting principle. The fundamental counting principle tells us that if there are m possible choices for one thing and n possible choices for another thing, I can simply multiply these together, m⋅n, to get the number of total possible choices for both of those things together. So, for my 3 shirts and 4 pairs of pants here, I can simply take 3 and multiply it by 4 to get my total of 12 possible outfits without having to actually come up with them and count them out.

So now that we've seen the fundamental counting principle, let's go ahead and work through some more examples together. Looking at this first example here, I see that a menu lists 4 appetizers and 6 entrees. And I want to know how many different meals with both an appetizer and an entree do we have to choose from. So, my first choice to make is of those 4 appetizers, so I have 4 possible choices there. Then for my second thing, I have 6 entrees to choose from, so I have 6 possible choices there. Now if I multiply those 2 together, 4 times 6, I get my number of total possible choices of a meal with an appetizer and an entree, and we're done here.

Now let's look at our second example. Here, we're asked to find how many possible outcomes there are if we flip a coin and then roll a 6-sided die. Well, something that you might notice here is that we're not really making a choice. If I flip a coin, there are 2 things that could happen: I could get heads or tails, but I'm not really choosing here. And that's okay because the fundamental counting principle also applies to outcomes of events. So if there are m possible outcomes of one event and n possible outcomes of another event, I can still multiply m⋅n to get the number of total possible outcomes of both events. So let's go ahead and do that here. When flipping a coin, we know that we can get heads or tails. There are only 2 possible outcomes here, so I'm going to take that 2 and then multiply it by the number of total possible outcomes of that second event. When I roll a 6-sided die, I could get any number 1 through 6, so I know that I have 6 total possible outcomes here. So multiplying those 2 together, 2 times 6, I get my total of 12 possible outcomes of both of these events together.

Let's take a look at one final example here. Here, we're asked how many different outfits can be made from 4 shirts, 5 pairs of pants, and 3 pairs of shoes. So the first choice I have to make is what shirt I'm going to wear out of these 4, so I know that I have 4 possible choices there. Then for my second thing, I have 5 pairs of pants, so I want to multiply 4 times 5. But since I have a third thing here, what exactly do I need to do? Well, the fundamental counting principle doesn't just apply to 2 events. It can also apply to any number of events. So if I have more than 2 things that I'm choosing from, I'm just going to continue to multiply by the number of options of each thing, and that will give us our total. So here, since I have 3 pairs of shoes, I'm going to continue to multiply, and I'm going to multiply 4 times 5 times 3, which gives me 60, my number of total possible options of those three things together. Now that we've seen the fundamental counting principle and know how to use it, let's get some more practice. Thanks for watching, and I'll see you in the next one.

How many possible outcomes are there if you roll 5 dice?

• A. 720
• B. 7,776
• C. 5
• D. 6

How many options are there for license plates with any three letters (A-Z) followed by any 3 numbers (0-9)?

• A. 260
• B. 2,340
• C. 11,232,000
• D. 17,576,000

Phone numbers are 10 digits long. How many possible phone numbers are there if the 1^st and 4^th numbers can't be 0?

• A. 10
• B. 90
• C. 8,100,000,000
• D. 10,000,000,000

Introduction to Permutations

Hey everyone! We just learned how to find the number of different ways we could wear different outfits on any given day, but what if, instead of a single day, we wanted to look at the week ahead and determine the different ways we could wear 5 different shirts over 5 different days? Well, the number of ways that we could choose to do this is actually something called permutations, which are just a way to arrange a number of things in a specific order with each thing occurring only once. Now that word permutations might make this sound like it's going to be complicated and a bit tricky, but you don't have to worry about that because here I'm going to walk you through exactly what permutations are and how we can calculate them using the fundamental counting principle and factorials, 2 things that we already know how to do. So let's go ahead and get started.

So when looking at our 5 different shirts that we want to figure out how to wear over 5 different days, we're going to use the fundamental counting principle and first look at the number of options that we have on that first day. So on Monday, I have all 5 different shirts to choose from, so I have 5 total options. But then on Monday, if I pick to wear this pink shirt, then on Tuesday I'm left with only 4 options. So if on Tuesday, I choose to wear this teal shirt over here, on Wednesday, I now only have 3 options left, then 2 on Thursday, and 1 on Friday. Now this 5×4×3×2×1, you might recognize as being a 5!, and you'd be right. The number of different ways we could wear 5 different shirts over 5 days is equal to 5 factorial.

But what if instead of 5 shirts, I actually had 8 different shirts to choose from? How would I choose that? Well, on that 1st day, I now have 8 total options because now I have 8 shirts to choose from. Then on that second day, I only have 7 and then 6 and then 5 and then 4. But this is not equal to 8 factorial. So if we can't always find permutations by just taking the factorial, how exactly are we going to calculate it? Well, permutations actually have a specific formula, this formula right here, that is based on the number of total options that we're given and how many things we're picking out of that total. So if I take this formula and apply it to my problem with 8 shirts over 5 days, I'm going to take the total number of options I'm faced with, in this case 8, and the factorial of that, and divide it by that total minus the number of things that I'm choosing. In this case, since I'm only choosing shirts for 5 days, my number here is 5, then the factorial of that. Now this simplifies to 8!÷3!, which is actually exactly what this here is.

So a couple of things that I want to point out about our equation here is that n is always going to be the largest number. So n! on the top, that n is always going to be your total. It's going to be the largest number that you're given in a problem. Then on the bottom, we're going to take that total and subtract our smaller number. Now in terms of notation, you may see this written as p n,r or p(n,r). Now this can be read as the number of permutations of n things taken r at a time. Now, that might sound a little bit overwhelming just because there are so many words going on, but let's go ahead and work through some examples together to make this a bit more clear.

So in our first problem here, we have a teacher that is choosing a line leader and a door holder from her class of 25 students. So, we want to know how many ways there are for the teacher to choose these positions. So using our permutations formula here, the first thing we want to do is identify our values for n and r. Now remember, n is always going to be your largest number here. So in this case, we have a total of 25 students, so this 25 represents our value of n. Now for r, we're looking at how many things we're choosing out of that total. So we're choosing a line leader and a door holder, so that tells us that our value for r is equal to 2 because we're choosing 2 positions out of our total of 25 different students. So in that permutations notation, I would write that as P 25,2 . And from here, we're just going to plug these values into our formula. So remember, our largest number, our n, is always going to come on the top there, so that I have 25! on the top. Then on the bottom, I take that 25 and I subtract that smaller number, in this case, 2, and I take the factorial of that. Now simplifying this, this gives me 25!÷23!. Now, you may be worried that we're now going to have to write all of these factorials out, but you don't have to worry about that because there's actually a much easier way to solve this by simply rewriting the numerator in order to cancel out the denominator. So here, since our numerator is 25! and our denominator is 23!, we want to rewrite 25! in order to cancel that 23!. So I can rewrite that 25! as 25×24×23! using what we know about factorials here. Then since this gets divided by 23!, that will end up canceling out and give me my final answer of 25×24, which is simply equal to 600. So there are 600 different ways that the teacher could choose a line leader and a door holder from their class of 25 different students.

Now looking at our second example here, we see that on our homework there are 10 fill-in-the-blank questions and a word bank of 14 words. If we can only use each word once, how many possible ways could we answer these 10 questions? So again, the first thing we that we want to do here is identify our values of n and r. So remember, n is always going to be your largest number. In this case, 14. This represents my total. Then r is my smaller number that I'm choosing from my total. In this case, I'm choosing 10 words out of my total of 14. So 10 represents my value for r. Now writing this in our permutations notations, this is written as P 14,10 . Now we're just left to plug everything into our formula here. So in this formula, we're going to take that 14! on the top, our largest number. And then on the bottom, we have 14−10!. Now this simplifies to 14!÷4!. And again here, we want to go ahead and rewrite the numerator to cancel the denominator. Now something that you might notice here is that we actually could just type this into our calculator if we know how to use the factorial button. And there actually is a specific function on your calculator to do permutations. But if you're not quite sure how to use that, we can always work this out by hand. So let's continue here and rewrite our numerator to cancel our denominator. Now this one is going to get a bit long, but that's okay. So let's go ahead and start. Now on that numerator, I can simplify this to 14×13×12 all the way down to 4!. So now in the denominator, that 4! will cancel with that 4! on the top, leaving me with all of this 14 counting down to 5, multiplying all of that along the way. Now if I type this in my calculator now, multiplying all of that, I'm going to get a value of 636,204,800. So these are all of the different ways that we could fill in these 10 questions. So it's safe to say that guessing is not going to be your best bet here. So now that we know what permutations are and how to calculate them, let's get some more practice. Let me know if you have any questions.

A student formed a club at their school. They have 13 members, and need to elect a president, vice president, and treasurer. How many ways are there to fill these officer positions?

• A. 2197
• B. 1716
• C. 13
• D. 6

Emily is organizing her closet. She has 15 shirts left to hang but has space in one section for 6 shirts. How many ways could she hang shirts in that section?

• A. 3,603,600
• B. 90
• C. 9
• D. 362,880

Evaluate the given expression. $_9P_4$

• A. 24
• B. 3,024
• C. 15,120
• D. 362,880

Permutations of Non-Distinct Objects

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.

You want to arrange the books on your bookshelf by color. How many different ways could you arrange 12 books if 4 of them have a blue cover, 3 are yellow, and 5 are white?

• A. 120
• B. 11,880
• C. 27,720
• D. 479,001,600

How many ways are there to arrange the letters in the word CALCULUS?

• A. 40,320
• B. 5,040
• C. 720
• D. 6

Permutations vs. Combinations

Hey, everyone. Up to this point, we've been working with the permutations of different objects. So when considering the permutations of two letters from a, b, and c, I would consider ab, but I would also consider ba. These are two different permutations because they're in two different orders. But now we're going to look at a new way to organize objects called a combination. Now distinguishing between permutations and combinations can be tricky, especially because these words are sometimes used interchangeably in everyday language. But here, I'm going to walk you through exactly what the mathematical difference is between these two things and how to distinguish between them by considering just one thing, whether or not the order of these objects matters. So let's go ahead and jump right in here.

Now in working with permutations, we saw that they were a different way to arrange objects in a particular order. So we know that when working with permutations, the order definitely does matter. So when writing all the permutations of two letters from a, b, and c, I would want to consider bc, but also cb, and ac, but also ca. Because these are in completely different orders, they represent different permutations. But when working with combinations, combinations are simply a way to group objects. So if I consider groups of 2 letters from a, b, and c, I might consider ab, bc, ac, but nothing else because when working with ab, this is the same group of letters as if I were to have ba. So when working with combinations the order does not matter at all.

Now this might seem a little abstract so let's go ahead and look at some different examples and determine whether we're working with a permutation or a combination. Looking at this first example here, I'm told that an ice cream shop has 32 different flavors, and we need to pick 2 flavors to blend into a milkshake. Now we want to know how many possible ways we can select these flavors. Now here we're just concerned with identifying whether this is a permutation or a combination, so we want to ask ourselves one question. Does the order of my objects matter? Well, if I'm blending up a milkshake, I might want to consider the outcome. So if I pick chocolate and vanilla and I blend that up, wouldn't that be the same milkshake as if I picked vanilla and then chocolate? Well, it would. So my outcome would be no different even if I changed the order. So since the order does not affect the outcome, here I'm simply working with a combination. The order does not matter.

Let's look at our next example. Here we're asked how many ways could a photographer line up the members of a family of 5? Well, if I were taking this photo, I might consider a couple of different things. I might consider the height of the people in it or the outfits that they're wearing. So if I put one person on the end in one of those photos, but then I move them to the other side in another photo, this would be a completely different photo. So if I change the order, that affects the outcome. So because the order affects the outcome, that tells me that order does matter, and here we're working with a permutation.

Let's look at one final example here. Here we're asked how many different teams of 4 people can be formed from a group of 9 people. Well, let's say that I pick 4 people out of this group of 9. Well, if I pick this person first instead, is that going to change what my team is? Well, it's not going to change it at all besides maybe starting a fight between teammates. So because the order does not have an impact on the outcome, here the order does not matter and we're working with a combination. So now that we know how to distinguish between permutations and combinations, let's keep going. Thanks for watching, and I'll see you in the next one.

Combinations

Hey, everyone. When finding the number of permutations of 2 letters out of these three letters, knowing that the order matters, whenever I list these out, I end up with 6 permutations. But when considering the combinations of 2 letters out of these three letters, knowing that the order does not matter, I only get 3 combinations. Now just seeing that these give me different numbers, you might realize that we're going to have to calculate these two things differently. But don't worry. We're not just going to have to learn a brand new equation for combinations because we're just going to take our permutations formula and alter it slightly in order to account for the fact that order doesn't matter in order to get our new combinations formula. So with that in mind, let's go ahead and get started.

Now, when working with our combinations, if we just plug in our values for n and r, we end up getting 3!1!, which is just equal to 6. Now, for our combinations formula, it looks almost identical to our permutations formula, but now we're just dividing by our factorial. So you'll see that I have my permutations formula right here, and I'm just going to add this extra factor of r! on the bottom here. So whenever I compute my combinations formula, plugging in my values for n and r, I end up getting 3!(1!×2!), which gives me my answer of 3.

Now with this formula, remember that n is always going to be your largest number because this represents your total and r is the number of things you're choosing out of that total. Now one way that you could remember this formula, seeing that we have an n and then an r, whenever you think of combinations, you can think n,n,r,r in order to remember the order that these variables come in.

Now something else that might seem familiar here is this combinations looks similar to our permutations, well, with the c with the n,r as subscripts or c with n,r in parentheses. Both of these can be read as the number of combinations of r things taken n at a time. Now that we know what our combinations formula is, let's get a bit more practice using it by working through some examples together.

So looking at this first example, we're told that an ice cream shop has 32 flavors and we need to pick 2 flavors to blend into a milkshake. We want to know how many possible ways we could do this. Now remember, this is a combinations problem because the order doesn't matter. A chocolate and vanilla milkshake is the same thing as a vanilla and chocolate milkshake. Now the first thing we want to do here is identify our values for n and r. Remember that n is always going to be the largest number, our total, So here my value for n is 32. Then this 2 right here represents my value for r. So plugging this into my combinations notation, I get C322, and then I'm just plugging these numbers into the formula. So I start with 32 factorial on the top, your largest number, and then I have 32 minus 2 factorial. And don't forget that extra factor of r! on the bottom there.

Now, simplifying this a bit more, I have 32 factorial on the top divided by 30 factorial times 2 factorial. Now here, since we have multiple factorials in that denominator there, remember, we can rewrite our numerator in order to cancel the highest factorial in the denominator. So here, we want to cancel this 30 factorial, So let's rewrite that numerator to cancel it. So 32 factorial can be rewritten as 32 times 31 times 30 factorial. And then in my denominator, I have 30 factorial times 2 factorial. Now those 30 factorials will cancel, leaving me on the top with 32 times 31. And then on the bottom, expanding that 2 factorial, that's just 2 times 1. Now, actually multiplying that out gives me 992 on the top divided by 2 on the bottom, giving me a final answer of 496 different possible ways we can select our flavors here.

Let's move on to our next example here. Here we're asked how many different teams of 4 people can be formed from a group of 9 people. So let's first identify our values of n and r. Now remember n is your highest number, so this 9 right here, and then r is our other number, so 4. Now writing this as a combination, we have C94, and we can plug in those values to our formula. So starting with n on the top, of course, we have 9 factorial divided by 9 minus 4 factorial times 4 factorial. Now simplifying this a bit further, we end up with 9 factorial over 5 factorial times 4 factorial.

Now, again, we want to cancel the highest factorial in the denominator here by rewriting that numerator. So here, I'm going to rewrite my numerator as 9 times 8 times 7 times 6 times 5 factorial. And then on the bottom, I have that 5 factorial and 4 factorial, but those 5 factorials are going to cancel. So on the top, I am left with 9 times 8 times 7 times 6. And on the bottom, I have that 4 factorial, which expanded out is 4 times 3 times 2 times 1. Now multiplying that out, I get a rather high number in my numerator here, 3,024, and then on the bottom, I get 24. Now actually dividing that gives me a final answer of 126 different teams of 4 people that can be formed from that 9. Now that we know how to calculate combinations, let's get some more practice. Thanks for watching and let me know if you have questions.

From a class of 28 students, in how many ways could a teacher select 4 students to lead the class discussion?

• A. 491,400
• B. 24
• C. 11,793,600
• D. 20,475

Evaluate the given expression. $_{11}C_7$

• A. 330
• B. 120
• C. 5,040
• D. 7,920

Here’s what students ask on this topic:

How do you calculate the number of possible outfits with multiple clothing items using the fundamental counting principle?

The fundamental counting principle states that if you have multiple choices for different items, you can find the total number of combinations by multiplying the number of choices for each item. For example, if you have 3 shirts and 4 pairs of pants, the total number of possible outfits is calculated as:

$3\times 4=12$

This means you have 12 different outfits to choose from. This principle can be extended to more items, such as adding shoes to the mix. If you have 3 pairs of shoes, the total number of outfits would be:

$3\times 4\times 3=36$

What is the difference between permutations and combinations?

The key difference between permutations and combinations is whether the order of the items matters. In permutations, the order matters. For example, the permutations of the letters A, B, and C taken two at a time are AB, BA, AC, CA, BC, and CB. In combinations, the order does not matter. For the same letters, the combinations taken two at a time are AB, AC, and BC. The formulas for permutations and combinations are:

Permutations: $P(n,r)=\frac{n!}{(n-r)!}$

Combinations: $C(n,r)=\frac{n!}{(n-r)!\times r!}$

How do you calculate permutations when the order matters?

To calculate permutations when the order matters, use the formula:

$P(n,r)=\frac{n!}{(n-r)!}$

Here, n is the total number of items, and r is the number of items to arrange. For example, if you have 8 shirts and want to choose 5 to wear over 5 days, the calculation is:

$\frac{8!}{(8-5)!}=\frac{8!}{3!}=6720$

How do you calculate combinations when the order does not matter?

To calculate combinations when the order does not matter, use the formula:

$C(n,r)=\frac{n!}{(n-r)!\times r!}$

Here, n is the total number of items, and r is the number of items to choose. For example, if you have 9 people and want to form a team of 4, the calculation is:

$\frac{9!}{(9-4)!\times 4!}=\frac{9!}{5!\times 4!}=126$

How do you calculate permutations with non-distinct objects?

To calculate permutations with non-distinct objects, use the formula:

$\frac{n!}{r!\times s!}$

Here, n is the total number of items, and r and s are the counts of each type of non-distinct item. For example, if you have the word 'banana' with 6 letters (3 a's, 2 n's, and 1 b), the calculation is:

$\frac{6!}{3!\times 2!\times 1!}=60$