Permutations
Sometimes you want to form sets of objects from a larger population of objects and sometimes the order of those objects matters! For instance, if you have a pile of 10 beads and you intend to make bracelets that use three different beads, but you consider each permutation (blue, red, purple and purple, blue, red) different, then you want to find how many permutations you can create.
The formula for permutations is $\dfrac{n!}{(n-r)!}$. If you are allowed to use a calculator, most scientific and graphing calculators also have a combination button, which usually looks like: $\boxed{_nP_r}$.
Before we get into the details of the formula and the button, let's talk about exactly what a permutation is.
There are a number of examples of combinations in the lesson Combinations. Here, we will discuss permutations.
Remember: There are almost always more permutations than combinations possible. Why is that? Within every combination (ABC) there are multiple permutations (BAC, BCA, CAB, CBA, ACB). When you are counting permutations, you are not just counting groupings, but the order of the objects within those groupings.
When might you want to count permutations?
- When you are trying to figure out schedules, order matters. So, if you have to choose 4 classes (out of 10 options) but want to figure out not how many different combinations of classes you could have (that would be a combinations problem) but what your schedule would be (ok, you have history, English, math, and Spanish, but is it history then English, then math, then Spanish? Or Spanish, then English, then... you get the point), you have a permutations question.
- When you are not just figuring out how many people can be on a team (or in a group) but what role they will play, you are counting permutations. So, if 20 people run for office and you want to see how many groups of officers you'll have, you want to count permutations. But if you consider a group different if Mary is president versus when Mary is treasurer, then you want to count permutations.
- When order just matters, then you want permutations. So, if you are figuring out what items to put on a sandwich or a pizza, order doesn't matter, lettuce can go on first or last and not change the sandwich. But, if you are arranging letters in acronyms, CAT is definitely different than ACT (though they would count as one combination).
Basically, if order matters, you want to count permutations. Often test questions will specifically say that order matters (that means they want you to count permutations) or that order does not matter (then, they want you to count combinations!).
Whether you use the formula or a calculator, you need to know some pieces of the puzzle.
First make sure that you want a permutation (that the order of the objects will make a difference).
Then figure out the total number of items that are being chosen for the permutations: n
Then figure out the number of items per permutation: r
Then you plug those numbers into the formula or into your calculator!
Example: You are trying to create a new acronym for a club. But, someone gave you the letters spelling DINER for free (they are really cool - you want to use them in a sign!). If you limit yourself to three-letter acronyms that use the letters in DINER, how many different acronyms do you have to choose from?
Order really matters here RED is different than DER; you want to find the number of permutations.
The number of possible letters is 5. $n=5$
The number of letters you want to choose is 3. $r=3$.
Let's try the two different ways of calculating the number of possible permutations:
| Use the formula | Use your calculator |
Important note: ! means factorial. Factorial means to multiply the number times every whole number lower than it. So, $4!=4\times3\times2\times1$ $$\require{cancel}\eqalign{\dfrac{n!}{(n-r)!}&=\text{Number of permutations}\\\dfrac{5!}{(5-3)!}&=\\\dfrac{5!}{2!}&=\dfrac{5\times4\times3\times2\times1}{2\times1)} \\\text{Reduce fractions before }&\text{ doing the multiplication!}\\\dfrac{5\times4\times3\times\cancel{2\times1}}{\cancel{2\times1}}&=\dfrac{60}{1}=60 \text{ Possible permutations}}$$ | Find the button that looks like $\boxed{_nP_r}$ on your calculator. It might be a "second" function that is written above one of your keys. If you have a graphing calculator, it might be in the Math/Prob menu. On most calculators, you will enter n, then press the button, then enter r. In this case, you will enter: $\boxed{5}\quad \boxed{_nP_r}\quad\boxed{3}\quad\boxed{\text{Enter}}$ Your calculator should yield: 60
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Can you believe that you can make 60 possible acronyms out of only 5 letters? (By the way, the number of combinations here is way smaller: 10)
Once you get used to working with factorials, the trickiest part of finding permutations is figuring out if you want combinations or permutations. Once you get that figured out, you figure out your set of options, the number of options you want and plug those values into the formula (or the calculator) and you'll get your number of permutations!
Practice Problems:
Permutations
Find the number of permuations:
- You have a set of 5. You are going to choose 3. How many permuations can you get?
- You have a set of 10. You are going to choose 6. How many permuations can you get?
- You have a set of 9. You are going to choose 2. How many permuations can you get?
- You have 8 different colored beads. You want to put 3 beads on a necklace. How many different necklaces can you make?
- You have a dozen crayons of different colors. You are making a tesselation that uses 4 different colors. How many different patterns can you make?
- You have 8 drawings. You have 4 frames. You are going to choose 3 pictures, put them in frames, and put them in different rooms in your house. How many different painting arrangements can you have?
- You are interviewing 7 people for three different jobs. How many differents staffs can you create?
- You have a set of letter 6 tiles. How many different 6 letter words can you make (if you don't care if the letters you put together actually make an English word!)?
Answer Key: