22
FACTORIALS
BY THE SYMBOL n! ("n factorial") we mean the product of consecutive numbers 1 through n.
| n! | = | 1· 2· 3· · · (n − 2)(n − 1)n |
| = | n(n − 1)(n − 2)· · · 3· 2· 1 | |
The order of the factors does not matter, whether backwards or forwards.
0! is defined as 1.
0! = 1
(The usefulness of this definition will become clear as we continue.)
Example. 6! = 1· 2· 3· 4· 5· 6 = 720
A convenient way to calculate this is to wait to multiply 2· 5 = 10. Then
3· 4· 6 = 12· 6 = 72
-- times 10 is 720.
Calculations with factorials are based on this fact:
Any factorial less than n! is a factor of n!.
Example 1. 6! is a factor of 10!. For,
| 10! | = | 1· 2· 3· 4· 5· 6· 7· 8· 9· 10 |
| = | 6!· 7· 8· 9· 10 | |
| Example 2. Evaluate | 8! 5! |
. |
| 8! 5! |
= | 1· 2· 3· 4· 5· 6· 7· 8 1· 2· 3· 4· 5 |
= | 6· 7· 8 | = | 6· 56 | = | 336 |
| Example 3. Evaluate | 8! 6! 2! |
. |
Solution. 6! is a factor of 8!, so it will cancel leaving 7· 8 in the
| numerator. We will have | 7· 8 1· 2 |
= 28. |
Problem 1. What factorial is each of these? (Consider what each symbol means.)
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a) 3!· 4 = 4! For, 3!· 4 = 1· 2· 3· 4 = 4!. b) 6!· 7· 8 = 8!
c) (n − 1)! n = n! d) (n − k − 1)! (n − k) = (n − k)!
| Problem 2. Evaluate | 10! 7! |
. |
| 10! 7! |
= | 7!· 8· 9· 10 7! |
= | 8· 9· 10 | = | 720 |
| Problem 3. Evaluate | 12! 10! 2! |
. |
| 12! 10! 2! |
= | 10!· 11· 12 10! 2! |
= | 11· 12 1· 2 |
= | 11· 6 = 66 |
| Example 4. Show: | 1 (n − 1)! |
= | n n! |
Solution. If we multiply both terms on the left by n, then
| 1 (n − 1)! |
= | n (n − 1)! n |
Now, (n − 1)! is the product up to the number just before n. Therefore, (n − 1)! times n itself is n!.
| 1 (n − 1)! |
= | n (n − 1)! n |
= | n n! |
Example 5. Show: n! k + n! (n − k + 1) = (n + 1)!
Solution. On the left-hand side, n! is a common factor:
| n! k + n! (n − k + 1) | = | n! (k + n − k + 1) |
| = | n! (n + 1), on canceling the k's, | |
| = | (n + 1)! | |
For, (n + 1) is the number after n.
Problem 4. Write out the steps that show how to transform the left-hand side into the right-hand side. Again, consider what each symbol means.
| a) | 1 3! |
= | 4 4! |
Multiply both terms on the left by 4:
| 1 3! |
= | 4 3!· 4 |
= | 4 4! |
| b) | 1 (k − 1)! |
= | k k! |
Multiply both terms on the left by k:
| 1 (k − 1)! |
= | k (k − 1)!· k |
= | k k! |
| c) | 12 12! |
= | 1 11! |
Since 12! = 11!· 12, divide both terms on the left by 12:
| 12 12! |
= | 12 11!· 12 |
= | 1 11! |
| d) | n − k (n − k)! |
= | 1 (n − k − 1)! |
Since (n − k)! = (n − k − 1)!· (n − k), divide both terms on the left by (n − k):
| n − k (n − k)! |
= | n − k (n − k − 1)!· (n − k) |
= | 1 (n − k − 1)! |
e) 5!· 4 + 5!· 2 = 6!
5! is a common factor:
5!· 4 + 5!· 2 = 5!(4 + 2) = 5!· 6 = 6!
Next Topic: Permutations and combinations
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