Let's look at two problems that involve factorials.
Problem 1:
If there are 8 different books on a shelf, how many ways are there to arrange those books?
Imagine placing all the books on the floor. One would have to choose between 8 different books to be placed first on the shelf. Once picked, there would be 7 books remaining that could also be picked for the second place on the shelf. Once that book is placed on a shelf, there would be 6 books to choose from, and so on.
Using the fundamental counting principal, the math would look like this.
There are 40,320 ways to arrange these 8 books.
Problem 2:
If there are 4 people who need to be seated at a circular table, how many different ways can this be done?
It appears this problem is the same as problem #1 above but it has less objects to arrange. We could list all the possibilities. We will list the people as person A, person B, person C, and person D.
The problem with this list, which lists all combinations of A, B, C, and D, it does not take in to consideration that there is a circular table. However, it is a great list for placing people in a line.
The difference is, there are situations listed there that are the same. Here are arrangements that are exactly the same, because of the circular situation.
Take any letter. Look to its left and right and you will see identical adjacent letters in each set. This means the only true unique situations we can gain is by looking down one of the columns from our large 24-set list above. So, there really are only 6 unique arrangements.
This means if we take one less than our number of people, we can simply use factorials.
There are 6 ways to arrange 4 people at a circular table.
Note: This means if we started with 8 people, we could calculate the total possible ways to arrange them around a table by calculating 7!, which is 5,040.
Kyden L. mathhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
1 Expert Answer
Abigail H. answered • 04/10/20
PA Certified 7-12 Math Tutor
Assuming that there cannot be a tie, here is how you can approach this problem.
There are five runners, so there are five different possibilities of who can come in first place.
Now, once someone comes in first place, there are only 4 runners left who can come in second place.
Likewise, there are only 3 runners left who can come in third place.
You want to then multiply 5*4*3 to represent the different combinations we can make for each place.
This gives us a total of 60 different possible ways that the runners can come in 1st, 2nd, and 3rd place, assuming no ties.
In a 10 man race, how many runners have the chance to finish first? answer 10. Now that 1 person is first, how many runners could be second? Well, there are still 9 runners left, and any one of hem could be second. Answer 9. How many can be third? Using the same idea as above, once one person is first and another is is second, that leaves 8 people still running and trying for third. So how many can be in third? Answer 8. Now you need to do the math
So there are 720 different ways
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