How to Simplify Factorials
Factorials - How to Simplify - How it Works - Video
Definitions
Definitions:
The symbol ! in math is not a exclamation mark. It tells us to multiply all whole numbers until we reach 1.
We have to definitions. The first one is what the sentence above tells us. Look at example 1 to see it in use.
The second one is stopping after one number. Look at example 2 to see it in use. This is one is helpful so we don't have to write each number which can be cumbersome.
Example 1
Example 1:
Here we have 4! or 4 factorial.
So we start at 4 and write down each whole number until we reach 1.
4 * 3 * 2 * 1 = 24
This means 4! = 24.
Example 2
Example 2:
Here we have 7!/5! or 7 factorial over 5 factorial.
On the left side we use definition 1 and on the right side we use definition 2.
So on the left side using definition 1, we have
(7 * 6 * 5 * 4 * 3 * 2 * 1) / (5 * 4 * 3 * 2 * 1)
We cancel all the numbers that are same on top and bottom and we have left with 7 * 6 = 42. So 7!/5! is 42.
So on the left side using definition 2, we have
(7 * 6 * 5!) / (5!)
Here we wrote our factorials and stopped when we have one in the numerator and one in the denominator. The reason is that mathematicians want to write the least possible.
Here we can cancel the 5! on top and on bottom. After that we are left with 7 * 6 = 42. So 7!/5! is 42.
Example 3
Example 3:
Here we have [4!9!] / [5!7!] or 4 factorial times 9 factorials all over 5 factorial times 7 factorial.
On the left side we use definition 1 and on the right side we use definition 2.
So on the left side using definition 1, we have
[(4 * 3 * 2 * 1) * (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)] /
[(5 * 4 * 3 * 2 * 1) * ( 7 * 6 * 5 * 4 * 3 * 2 * 1)
We cancel all the numbers that are same on top and bottom and we have left with 9 * 8 on top which is 42, and 5 on the bottom. So [4!9!] / [5!7!] is 72/5.
So on the left side using definition 2, we have
[(4!) * (9 * 8 * 7! ] / [(5 * 4!) * (7!)]
Here we wrote our factorials and stopped when we have one in the numerator and one in the denominator. The reason is that mathematicians want to write the least possible.
Here we rewrote 9! as 9 * 8 * 7! because we have 7! on the bottom, and we rewrote 5! as 5 * 4! because we have 4! on top.
After cancelling the factorials that are the same since one is top and one is on bottom, we are left with 9 * 8 on top, which is 72, and 5 on the bottom. So [4!9!] / [5!7!] is 72/5.
Example 4
Example 4:
Here we have factorials with variables. This time we are only going to use definition 2.
Here we have [(n + 2)!] / [n!].
Since the term in the numerator, n + 2, has a higher value when you substitute any number than n, we are to change (n + 2)! by following definition 2.
So (n + 2)! becomes (n + 2) * (n + 1)! Right now we can simplify yet because we don't have matching terms in the numerator and denominator.
Let's use the definition one more time. (n + 2) * (n + 1) * (n - 0)! or (n + 2) * (n + 1) * (n)! Now we can simplify.
So [(n + 2) * (n + 1) * (n)!] / (n)! is (n + 2) * (n + 1).
Example 5
Example 5:
Here we have factorials with variables. This time we are only going to use definition 2.
Here we have [(4n + 2)!] / [(4n-1)!].
Since the term in the numerator, 4n + 2, has a higher value when you substitute any number than 4n - 1, we are to change (4n + 2)! by following definition 2.
So (4n + 2)! becomes (4n + 2) * (4n + 1)! Right now we can simplify yet because we don't have matching terms in the numerator and denominator.
Let's use the definition one more time. (4n + 2) * (4n + 1) * (4n - 0)! or (4n + 2) * (4n + 1) * (4n)! We still can't simplify.
So let's use the definition one more time. (4n + 2) * (4n + 1) * (4n) * (4n - 1)! Now we can simplify.
So [(4n + 2) * (4n + 1) * (4n) * (4n - 1)!] / (4n - 1)! is (4n + 2) * (4n + 1) * (4n).