Check this out.
You have three types of fruit (peach, watermelon, apple.) You need to pick four fruits. The order is irrelevant, and you can have more than one of each kind of fruit. You don't need to use all the fruits. What are all the possible combinations?
The formula is (6!) / (4! * 2!) = 15. There are fifteen combinations. Now, you might wonder, where the heck does that formula come from? After all, you're only picking four elements. Why is the numerator six factorial?
Well, you can think of it this way.
You're picking out four fruits. Picture the four empty slots as Xs:
X X X X
Now, there are three categories of fruit available, right? You can break the four slots into three categories by placing two (2) dividers.
x | x x | x
With me? All make sense? We could say 'left slot is peach, middle slot is watermelon, right slot is apple', like so:
peach | watermelon, watermelon | apple
So, in fact you can define any configuration of fruits in the four slots by moving the two dividers. A four-peach configuration would be:
x x x x ||
See? Four peaches in the left slot, nothing in the watermelon or apple slots.
So we're actually manipulating the positions of six elements: the four fruit slots and the two dividers.
Thus, 6! as the numerator.
The denominator is the factorial of the number of slots (4!) multiplied by the factorial of the number of dividers.
Thus, (6!)/(4!*2!)
You can generalize this problem to any combination problem where you're fitting N types of things into X slots (with repetition allowed) as
(X + n-1)!/(X! * (n-1)!)
Isn't that sweet?
