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# ... in which I demonstrate how much I've forgotten since undergrad

OK math geeks.

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?

MY SOLUTION:

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.

MY QUESTION:

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. ? Anyone?
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