How do you distinguish permutations from combinations?
short answers for big questions
This is a very common doubt felt by most students who are starting their study of probabilities. Let´s start by distinguish, by means of some examples, between full and simple permutation. Then we will explain what are combinations.
Full permutation
A mobile PIN is composed of 4 digits. How many distinct possibilities are there? Answer: as the digits can be repeated and the order is a focus of interest, we will have a full permutation. Since there are 10 different possibilities for each digit, we calculate as follows: `10^4=10000`.
Simple permutation
Coming back to the mobile PIN, let’s us suppose now that we do not want to use repeated figures. In this case, how many possibilities are there? Answer: as the digits cannot be repeated and the order is a focus of interest, we will have a simple permutation and is calculated the following way: `text()^10P_4 = (10!)/((10-4)!)=10xx9xx8xx7=5040`.
Combinations
I am going on holiday and I want to take 3 or 7 books that I have just bought. How many choices do I have? Answer: as the books cannot be repeated and the order is not a matter of interest, we will have a combination. It is calculated like this: `text()^7C_3 = (7!)/(3!(7-3)!)=35`
In conclusion, most of these exercises require a simple analysis to see if there is a matter of interest so as to decide which the best way to achieve the right result is. Here it is a table that will summarize the subject.
| Full permutation | Simple permutation | Combinations | |
|---|---|---|---|
| Does the order have any influence? | Yes | Yes | No |
| Is it possible to repeat elements? | Yes | No | No |
| Calculation | `text()^nA_p^' = n^p` | `text()^nA_p = (n!)/((n-p)!)` | `text()^nC_p = (n!)/(p!(n-p)!)` |
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