I’ve always confused “permutation” and “combination” — which one’s which?

Here’s an easy way to remember: **permutation sounds complicated**, doesn’t it? And it is. With permutations, every little detail matters. Alice, Bob and Charlie is different from Charlie, Bob and Alice (insert your friends’ names here).

Combinations, on the other hand, are pretty easy going. The details don’t matter. Alice, Bob and Charlie is the same as Charlie, Bob and Alice.

**Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter).**

A joke: A "combination lock" should really be called a "permutation lock". The order you put the numbers in matters. (A true "combination lock" would accept both 10-17-23 and 23-17-10 as correct.)

## Permutations: The hairy details

Let’s start with permutations, or **all possible ways** of doing something. We’re using the fancy-pants term “permutation”, so we’re going to care about every last detail, including the order of each item. Let’s say we have 8 people:

```
1: Alice
2: Bob
3: Charlie
4: David
5: Eve
6: Frank
7: George
8: Horatio
```

How many ways can we award a 1st, 2nd and 3rd place prize among eight contestants? (Gold / Silver / Bronze)

We’re going to use permutations since the order we hand out these medals matters. Here’s how it breaks down:

- Gold medal: 8 choices: A B C D E F G H (Clever how I made the names match up with letters, eh?). Let’s say A wins the Gold.
- Silver medal: 7 choices: B C D E F G H. Let’s say B wins the silver.
- Bronze medal: 6 choices: C D E F G H. Let’s say… C wins the bronze.

We picked certain people to win, but the details don’t matter: we had 8 choices at first, then 7, then 6. The total number of options was 8 · 7 · 6 = 336.

Let’s look at the details. We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.

We know the factorial is:

Unfortunately, that does too much! We only want 8 · 7 · 6. How can we “stop” the factorial at 5?

This is where permutations get cool: notice how we want to get rid of 5 · 4 · 3 · 2 · 1. What’s another name for this? 5 factorial!

So, if we do 8!/5! we get:

And why did we use the number 5? Because it was left over after we picked 3 medals from 8. So, a better way to write this would be:

where 8!/(8-3)! is just a fancy way of saying “Use the first 3 numbers of 8!”. If we have **n** items total and want to pick **k** in a certain order, we get:

And this is the fancy permutation formula: You have **n** items and want to find the number of ways **k** items can be ordered:

## Combinations, Ho!

Combinations are easy going. Order doesn’t matter. You can mix it up and it looks the same. Let’s say I’m a cheapskate and can’t afford separate Gold, Silver and Bronze medals. In fact, I can only afford empty tin cans.

How many ways can I give 3 tin cans to 8 people?

Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re equally disappointed.

This raises an interesting point — we’ve got some redundancies here. Alice Bob Charlie = Charlie Bob Alice. For a moment, let’s just figure out how many ways we can rearrange 3 people.

Well, we have 3 choices for the first person, 2 for the second, and only 1 for the last. So we have 3 · 2 · 1 ways to re-arrange 3 people.

Wait a minute… this is looking a bit like a permutation! You tricked me!

Indeed I did. If you have N people and you want to know how many arrangements there are for **all** of them, it’s just N factorial or N!

So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. If we want to figure out how many combinations we have, we just **create all the permutations and divide by all the redundancies**. In our case, we get 336 permutations (from above), and we divide by the 6 redundancies for each permutation and get 336/6 = 56.

The general formula is

which means “Find all the ways to pick k people from n, and divide by the k! variants”. Writing this out, we get our **combination formula**, or the number of ways to combine k items from a set of n:

## A few examples

Here’s a few examples of combinations (order doesn’t matter) from permutations (order matters).

Combination: Picking a team of 3 people from a group of 10. C(10,3) = 10!/(7! · 3!) = 10 · 9 · 8 / (3 · 2 · 1) = 120.

Permutation: Picking a President, VP and Waterboy from a group of 10. P(10,3) = 10!/7! = 10 · 9 · 8 = 720.

Combination: Choosing 3 desserts from a menu of 10. C(10,3) = 120.

Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P(10,3) = 720.

Don’t memorize the formulas, understand why they work. **Combinations sound simpler than permutations, and they are. You have fewer combinations than permutations.**

## Leave a Reply

1358 Comments on "Easy Permutations and Combinations"

Thanks alot! This was actally a better explanation then my teacher could give us =]

Yeah nd that too in a kinda fun manner….as if explaining a kid. Great work!

Agreed – this article helped me understand it WAY better.

yea, same!

Conclusion

With the above discussion, it is clear that permutation and combination are different terms, which are used in mathematics, statistics, research and our day to day life. A point to remember, regarding these two concepts is that, for a given set of objects, permutation will always be higher than its combinations.

what r u talking about?

hallelujiah!

[…] (Brush up on combinations and permuations if you like). […]

Good explaination

can anyone show me how many combinations there are using only 1, 2, and 3 across and 1 to 6 down.and 3 down, in other words its a 3 x 6 grid. Please do not give me the formula, I just need to see something like this, 1x2x3-1x3x2-2x3x1-2x1x3-3x1x2-3x2xhope it makes sense

Hello,

If you make the chart, you would then have to go through and cross out anything duplicates (1x2x3, 1x3x2, 2x3x1, 3x2x1, 3x1x2, 2x1x3) because the represent the same thing. There is not easy way to get ride of the duplicates when making a chart.

its complicated

how to do 6c3

6c3= 6!/3!(6-3)! => 6!/3!(3!) => 20

20

6!/3!(6-3)!

=6!/3!×3!

=6×5×4×3!/3!×3!

=6×5×4/3!

=6×5×4/3×2×1

Simplify then u will get..

=20 ans

6c3=6p3/3!

=6!/(6-3)!3!

=(6*5*4*3!)/(3!*3!)

=(6*5*4)/(3*2)

=20

6c3 is 6!/3!3!

6x5x4x3x2x1/3x2x1x3x2x1

=20

6!/3!*3!

6c3 can be written as;

(6!)/(6-3)!(3)!

= 6*5*4*3!/3!*3*2*1 (3! can cancel out), hence you have

=120/6

=20, I hope that answers your question

6*5*4*3*2*1/6-3!*3!=6*5*4*3*2*1/3*2*1*3*2*1*=720/36=20

It is 6•5•4/3•2•=20

it’s 6p3/3p3 or (6*5*4)/3*2*1) which = 20.

20and 3

6!/((6-3)!*3!)

20

in your calculation, press 10 and press C (Combination) of and then press 3

120

I HAVE UNDERSTOOD THE DIFFERENCE TODAY AFTER A GAP OF 38 YEARS

Awesome! Glad you found it useful :)

Hi David, yep, you got the formula right. The number of permutations (ways to order 22 people of 40) is:

40 * 39 * 38 … * 24 * 23 * 22 * 21 * 20 * 19 = 40! / 18!

[Be careful of off-by-one errors, I had a mistake at first. 40 to 19 is 22 people (just like 40 to 39 is 2 people, even though 40-39 is 1)]

And the number of ways to re-arrange 22 people = 22!

So we divide the first by the second and get

40!/18!(22!) = 113,380,261,800

113 billion does seem huge, but there’s a lot of multiplications happening. There’s 56 ways to pick 3 people from 8, which seems pretty large as well.

It’s one of those things where human beings (all of us!) aren’t great at intuitively estimating the impact of exponential growth. The birthday paradox and the effect of compound interest are other examples of this. I think it’s because we don’t encounter such mind-boggling growth or large numbers in a way we can really experience (at a certain point, millions, billions, and trillions become “a lot”, even though a trillion is a *million* times bigger than a million).

This helped with my question… sort of

this is an awesome site!

Hi D, when you buy multiple tickets you would add up the chances. So 10 tickets would be 1/13,000,000 + 1/13,000,000 + 1/13,000,000 … = 10 / 13,000,000

So buying multiple tickets would increase your chances for that particular lottery. If you somehow bought half of the available tickets, you’d have a 50-50 chance. And if you bought all of the tickets you’d win :).

Did Cooper win the lottery after all?

Thanks a million! It makes sense now!

Thanks for the encouragement Aaron! Once I have enough posts I would love to turn it into a book :)

finally this makes sense

This still makes 0 sense

I understood the permutations. But then I got confused at the combinations (I think I got confused because they used the combinations for the same example as the permutations)

If you pick 3 people from a group of 10 (Example 1 in the *A few examples* paragraph.) wouldn’t you do 10 x 9 x 8 like in the permutations paragraph, but then because unlike a permutation, a combination doesn’t care if you mix up those 3 numbers so you would have to divide 3! from 10 x 9 x 8.

10 x 9 x 8 = 720

3! = 3 x 2 x 1 = 6

720/6 = 120

That is what the author got, and I understand that.

If this helps, my teacher explained it using the word GLOSS. IF we were to find the number of ways we could mix up its letters, we would do 5! because there are 5 letters in the word. You put one of the letters (L for instance) in first spot, which leaves 4 letters left (G, O, S, S). You get another letter (say it’s S) in second spot which leaves 3 letters left (G, O, S). And so on until you have 0 letters left. So you started off with 5 letters, and went down to 0, which is why you do 5 x 4 x 3 x 2 x 1 or 5! But there are two S’s and it does not matter where you put either S. SOLGS is the same as SOLGS, right?! I swapped the S’s but you don’t notice. Now, because there are two S’s, we have to divide 5! (possible combinations) by 2! (the 2 S’s that are interchangeable)

5! = 5 x 4 x 3 x 2 x 1 = 120

2! = 2 x 1 = 2

120/2 = 60

Therefore there are 60 possible combinations to arrange the letters G, L, O, S, S.

If we use the word EXCEED, we would do 6! (because there are 6 letters) divided by 3! (3 E’s) which is:

6! = 6 x 5 x 4 x 3 x 2 x 1 = 720

3! = 3 x 2 x 1 = 6

720/6 = 120

For EXCEED there are 120 ways we could arrange those letters.

Mind you, this was only for combinations.

If we did permutations…

For the word GLOSS, it would matter where we put the S’s and they would NOT be interchangeable. Then it would just be 5!, which is 120.

For EXCEED, the E’s would not be interchangeable either, so it would be 6!, which is 720.

I hope this helped you understand Permutations and Combinations better, and if not, writing this helped me understand it better, so thank you!! :)

Please explain in how many way we can arrange 5men and 5 women so the 2men can not come together………

I know the answer 2880…but please explain how….

5 women can be seated in a circular manner such that there is a vacant between them

(5-4)! since its circular you subtract 1

Now 5 men can fit in the remaining seats with no two adjacent men

4!*5!

=2880

that was very helpful thankyou

Hi Safa, for that question it helps to take it one step at a time.

In a test with only 1 question, how many ways can you be wrong? 3. (Suppose the right answer is D… you could answer A, B or C).

Now how about 2 questions? Well, you have 3 ways to get the first question wrong, and another 3 ways to get the second one wrong. So the total is 3 * 3 = 9. (Let’s say the right answer is D and D. Then AA AB AB BA BB BC CA CB CC are all wrong).

Similarly, if you have 3 questions, then there is 3 * 3 * 3 = 3^3 = 27 ways to get all answers wrong. (You can write it out but will take a while: AAA AAB AAC ABA ABB ABC ACA ACB ACC… you get the idea :) )

Hope this makes sense,

-Kalid

hi!!! where can i find the solutions to the problems stated above???

Hello

Hi Pras, thanks for the comment!

@Sahil: It’s a good question, but I want to make sure I’m not doing someone’s homework for them :).

In general, it’s easier to find the chance of “zero matches” and subtract this from 1, vs. finding the chance for 1 match.

So let’s find the chance for zero matches. Imagine picking your first shoe, A: nothing special here, you aren’t going get the match on a single shoe.

You pick shoe B: You have a 18/19 chance of getting zero matches (only A’s partner would match, of the 19).

You pick shoe C: You have a 16/18 chance of zero matches (only A and B’s partner would match, of the remaining 18).

You pick shoe D: You have a 14/17 chance of not getting any matches (only A and B and C’s partner would match, of the remaining 17).

If you multiply these chances you get the total chance for zero matches. Subtract from zero to get the chance for any match. At least I think that’s how it goes :)

No. Of ways a pair can be picked from 10 – 10c1

No. Of ways 2shoes (any) can be picked from 18(remaining) – 18c2

No. Of ways picking atleast one pair in 4 out of 20 = 10c1*18c2

I hope it’s crct, pls correct me if I’m wrong

from the first 4 odd numbers (1,3,5 and 7) and the following 4 even numbers (2,46, and 8), how many groups of 4 digit numbers can be formed consisting of:

a. 2 odd numbers and 2 even numbers?

b. all odd numbers?

c. 1 odd number and 3 even numbers?

Great Stuff! You should write a book!

it was really useful dude!!!

thnx a lot!!!

@Kalid

You sure are doing a great work here making things that look really hard/challenging/tasking quite easy.

Thanks

Suppose you have 13 football matches and each game has 3 chances win, lose & Draw. How can you get a chances of all possible outcomes?

Dear Kalid,

I’ll say you are really a blessing to me. I so much value the way you make difficult things turn so easy. I am really so so happy for stumbling upon your site today after so much search for better explanations on different topics I find so confusing.

I’m one of the many that loves maths a lot but also finds it so frustrating most of the times, especially when the exercises are so twisted than what was used in explaining the topic.

I have two sets of exams to write on maths in the next 2 weeks and have been facing so much difficulties in some topics.

I believe I have reached a point where I can grab a sit and learn comfortable and as much as I can before my test.

Thanks a million bro!!!

@O. Opara: Really glad it helped!

Well written!

I am a teacher and I like the way you explained it so I put a link to this site on my students assignments to read as an introduction to the idea before we go into the more complicated problems in class.

Nyz 1

well explained!!

Thanks dude this helped me a lot!

I’m having a stupid moment. I have a problem: how many combinations exist when one needs to select a team of 22 players from a squad of 40 players?

IS this 40!/22!(18!) = 113,380,261,800?

It seems a rediculaously large number! Please help me!

So what!! These are the number of ways .

No problem, glad it helped!

Worst planning

Brock, bad comment

if a train has 18 cars , and 3types of cargo must be transported, how many ways can the 3 types be transported if one type of cargo can at most occupy only ten cars per train?