|
| 3! | = | 3×2×1 |
| 10! | = | 10×9×8×7×6×5×4×3×2×1 |
| n! | = | n × (n-1) × ... × 1 |
| 0! | = | 1 |
| n! | = |
| k |
| Q | What about zero factorial? (i.e. 0!) |
| A | There's a standard exception to the above rule, which says that 0! = 1. |
Suppose you have three objects (lets call them A, B, and C), which need to be arranged in a certain order (going left to right, for example). How many ways can you arrange the three objects?
How would you even work something like that out?
Lets see: We have three choices for which object goes first, then we have two objects left; either of which could be placed second, then one object left which gets placed at the end.
So the number of ways of arranging the three objects is equal to 3×2×1 = 6 different ways. Hey, that's the same as writing 3!. Each different arrangement is called a "permutation" of A, B, and C.
What if you have more objects? Well, this can be extended to any number of objects. If you have n objects, each of which can be placed once, and all of which must be placed in a specific order; then the number of permutations is simply n!.
What if you have zero objects to arrange? Well, there's exactly one way to arrange zero objects - that is not to arrange them at all! So in a way, the rule that 0! = 1 makes sense.
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