Given four unique items, how many different combinations can be made?
Here's the "word-problem" version of it:
I have four decorative house number tiles. They're all different numbers. I want a different style of house number, and I thought maybe I could advertise on craigslist or something to see if someone else with the same house numerals might like to trade (instead of going to buy some).
So I got me to thinkin'. How many unique addresses can be formed with my four numerals?
x = 4!/(4-1)! + 4!/(4-2)! + 4!/(4-3)! + 4!/(4-4)!
x = 24/6 + 24/2 + 24/1 + 24/1
x = 4 + 12 + 24 + 24
x = 64
I hammered it out by hand and verified the solution.
Makes sense when you think about it. The number of one-numeral permutations is four. For example, if my address is 7492, the one-numeral permutations are 7, 4, 9, and 2.
The number of two-numeral permutations follows easily: there are three for each one-numeral permutation. For 7, they are 72, 74, and 79. Four numerals, three permutations each = 12 two-number permutations.
Three numeral permutations: there are two for each two-numeral permutation. For 72, there are 724 and 729. Twelve two-numeral permutations, two permutations each = 24 three-number permutations.
Finally, each three-numeral permutation has only one possibility to be made into a four-numeral permutation: add the last remaining number. For 724, the only available four-numeral permutation is 7249. 24 three-number permutations, one permutation each = 24 four-number permutations.
For a total of 64 different combinations that can be made from a set of four different house number tiles.
The mathematical formula I used above seems clunky and inelegant, and it seems to me that there should be a cleaner way to set it up. sporkly? Anyone?