Introduction
Combinations and permutations are common throughout Mathematics and statistics, hence are a useful concept that us Data Scientists should know.
In this post, I want to discuss the difference between the two, difference within the two and also how one would calculate them for some given data.
Overview and Simple Example
The main thing that differentiates between permutations and combinations is that for the former order does matter but it doesn’t for the latter.
For example, let’s say we have three different coloured balls red, green and blue and we want to put them in an arbitrary order such as:
1: RED
2: GREEN
3: BLUE
The combination of these three balls is 1 as each ordering will contain the same three combination of balls. However, there are 6 permutations as we can have:
1: RED 1: RED 1: GREEN 1: GREEN 1: BLUE. 1: BLUE
2: GREEN 2: BLUE 2: RED. 2: BLUE 2: RED 2: GREEN
3: BLUE 3: GREEN 3: BLUE. 3: RED 3: GREEN 3: RED
Now you have a basic understanding of what combinations and permutations mean, let’s get more into the theoretical details!
Permutations
Like we said, for permutations order is important and we want all the possible ways/lists of ordering something.
There are actually two types of permutations:
Let’s run through both!
With Repetition
This one is pretty intuitive to explain. For example, given a padlock which has options for four digits that range from 0–9.
What are the code permutations for this padlock?
Well the first digit can have 10 values, the second digit can have 10 values, the third digit can have 10 values and the final fourth digit can also have 10 values. So, there are 10 x 10 x 10 x 10 = 10,000 permutations!
Mathematically, the formula for permutations with repetition is:

Where we have n choices at each r stage.
Without Repetition
Let’s go back to our ball analogy where we want to put three coloured balls red, green and blue into an arbitrary order.
How many permutations are there for three different coloured balls?
The first ball can go in any of the three spots, so it has 3 options. The second ball can then fill any of the remaining two spots, so has 2 options. Finally, the last ball only has one spot, so 1 option.
In that process each ball could only be used once, hence there was no repetition and our options decreased at each choice. In this case, we had 3 options, then 2 and then 1. Mathematically we had:

The exclamation mark is the factorial function. For example, n! is the product of all integers from 1 to n.
Now lets reframe the problem a bit.
How many permutations are there of selecting two of the three balls available?
Well at first I have 3 choices, then in my second pick I have 2 choices. We also have 1 ball left over, but we only wanted 2 choices!
To account for this we simply divide by the permutations left over. In our case this is luckily just 1!:

Let’s go through a better example to make this concept more concrete.
What are the permutations of selecting four cards from a normal deck of cards?
The first card we pick is out of 52 options, second one 51, third is 50, fourth is 49 and so on. As we only want the permutations from the first 4 cards, we have to divide by the remaining permutations (52 – 4 = 48):

That is a lot!
An alternative simple way would just be to calculate the product of 52, 51, 50 and 49.
In general, the formula for permutations without repetition is given by:

- n is the total number of things
- r is the number of things we choose
One can use the formula to verify all the example problems we went through above.
Combinations
For combinations order doesn’t matter, so (1, 2) = (2, 1).
Similarly, to permutations there are two types of combinations:
Let’s go through both scenarios.
Without Repetition
Lets once again return to our coloured ball scenario where we choose two balls out of the three which have colours red, blue and green.
How many different combinations of two different balls can we select from the three available?
Well the permutations of this problem was 6, but this includes ordering. To account for the ordering, we simply divide by the number of permutations of the two elements:

Which makes sense as we can have: (red, blue), (blue, green) and (red,green).
So to get the combinations, we calculate the permutations and divide by the permutations of the number of things we selected.
In general, the formula for combinations without repetition is given by:

- n is the total number of things
- r is the number of things we choose
This is often expressed as ‘n choose r’ using the binomial coefficient.

One can use the formula above to verify the results to the examples we discussed above.
With Repetition
This is the hardest one to grasp out of them all.
For this example, we will return to our almighty three different coloured balls (red, green and blue) scenario and ask:
How many combinations (with repetition) are there when we select two balls from a set of three different balls?
As we are allowed to repeat balls we can have combinations such as: (blue, blue), (red, red) and (green, green). So when we pick one ball, it is as if that same ball magically spawns back into our choices for the next ball we can choose.
These 3 new combinations are an addition to the number of combinations without repetition we calculated above, which was 3. Therefore, the total combinations with repetition for this question is 6.
The formula for combinations with repetition is:

- n is the total number of things
- r is the number of things we choose
The full derivation for this general formula is quite long arduous, therefore I have linked a full derivation here for the interested reader!
Conclusion
In this article we have explored the difference and mathematics behind combinations and permutations. The main thing to remember is that in permutations the order does not matter but it does for combinations!
Another Thing!
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