Basic Algebra and Geometry Made a Bit Easier Lesson Plans:
A Guide for Tutors, Parents, and Homeschoolers
Available exclusively on Amazon.com for $4.50. Please click Book Info below for details.
Home | Book Info | Math Lessons | Ask a Math Question | Site Info | Contact Info | Tutoring Info
Lesson 139: Intro to Permutations
In this very important lesson we'll take the counting principle to the next step, and learn how to solve questions that involve the numbers of ways that items can be arranged or chosen.
Here is very typical question: How many different arrangements of the the letters MATH can be made? There are 4 ways that we can choose the first letter. Once that letter is chosen, we cannot use it anymore. Now there are only 3 ways to choose the second letter. You can see that there are two ways to choose the third letter, and then only one letter is left, so there is just that one way to choose the last letter.
To find the total number of arrangements, we can multiply these numbers according to the counting principle. We do 4 x 3 x 2 x 1 = 24. There are 24 different ways that the letters can be arranged. We call these arrangements permutations. A permutation is a grouping in which order matters.
We have a special way of abbreviating 4 x 3 x 2 x 1. We can write 4!, which is read as "four factorial." The exclamation point doesn't mean that we're excited about the number 4. It means that we start with 4, and then keep multiplying by one number lower, then one number still lower, until we've reached 1. This is convenient when we're taking the factorial of a very large number. Most advanced calculators have a factorial function that does the computation for you. So in this example, we can say that there are 4! possible permutations of the letters MATH. Our general formula is that n objects can be arranged in n! different ways.
Very often we are asked to start with n objects, but only arrange a certain number of them. For example, from a group of 7 students, you must arrange 3 students in a row, where order matters. Here our formula is a bit different, since we're not ordering the entire group. There are 7 ways to pick the first student. With that student already chosen, now there are 6 ways to pick the second student, and then 5 ways to pick the third student. According to the counting principle, we have 7 x 6 x 5 = 210 possible permutations. Notice that we did not use factorial, since we weren't involving every student.
We do, however, have a special formula that we can use for
problems like this. The formula is: To arrange r objects
from a group of n total objects, the number of possible permutations
is: nPr = n! / (n-r)!.
Let's look at what all this means. The P stands for permutation. The n is the total number of objects in question. In the example above, that was 7. The r is how many we will actually be ordering. Above, it was 3. Using our formula to solve the question about, we would write 7P3 = 7! / (7-3)!. Of course 7-3 is 4, so now we have 7! / 4!.
Let's look at what we have here. 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1. 4! = 4 x 3 x 2 x 1. Can you see what cancels out? We have 4 x 3 x 2 x 1 in both the numerator and denominator. They just cancel out. All that is left is 7 x 6 x 5, which is what we actually multiplied in our problem above.
The formula can be tricky to memorize, but you need to. If you forget it, it is not hard to recreate. Just remember that permutations of n objects in general involve n!. But if we only want to arrange r of them, then we need to divide n! by the difference of n and r, so that we'll only be left with r terms in the numerator.
This might be easier to see with a more extreme example. Let's computer 97P2. Here we have 97 objects, and we only want to arrange two of them. We can start with 97!, but then we must divide by 95!, so that all that remains is 97 x 96.
Many permutation problems involve a special condition or restriction. In these cases, we cannot use our formula. We just have to figure out how many possible objects can be placed in each position, and then use the counting principle to multiply the results.
For example, how many arrangements of letters ABCDE can be made, if the last letter must be a vowel, and if letters cannot be repeated? Here we must start with the rightmost position. Only 2 letters can go there: A and E. Now that we have chosen a letter, it is gone. Let's start on the left. How many letters can go in the first slot? There are 4 left. Now two letters are gone. We can choose from the 3 that are left for the second slot, 2 for the third slot, and then only 1 remains for the fourth slot. What we have is 4 x 3 x 2 x 1 x 2. Remember that the last slot was for our two possible vowels. The answer is 48.
Let's do that same problem, but allowing letters to be repeated. Still only two possible letters can go in the rightmost position. But now how many can go in the first slot? Any of the 5. How many can go in the second slot? Again, any of the 5, since we're allowing repetition. We end up with 5 x 5 x 5 x 5 x 2. The only restriction was on the rightmost place. The answer is 1250. You can see that when you allow repetition, there are many more possible arrangements.
There are many different variants of problems of this type. What you must do is always read the problem very carefully, and make sure that you understand all of the restrictions and conditions. Always work with those first. If it will help you, write out some sample permutations by hand, just to make sure that you're understanding how the restrictions and conditions work. These problems take practice. Work on as many as you can, and try to make up your own variations of problems that you are given.
Remember that you can ask a math question if you have additional questions about a topic, or you can contact me if you have any comments or suggestions for this site.