Question

A hardware store sells single digits to be used for house numbers. There are five 5s, four 4s, three 3s and two 2s available. From this selection of digits, a customer is able to purchase her three-digit house number. How many possible three-digit house numbers could this customer make?

Answer 1

Since there are no restrictions on the three-digit numbers (for example no repetitions), we actually don't care about the values of the digits. All we need to know is that there are

$5+4+3+2 = 14$

possible digits, and that we have to extract a triplet from here.

For problems like this we have the binomial coefficient, defined as

$\displaystyle \binom{n}{k} = \dfrac{n!}{k!(n-k)!},\qquad 0\leq k \leq n$

This number tells you how many subsets of k elements you can extract from a set of n elements. So, in your case, you want to compute

$\displaystyle \binom{14}{3} = \dfrac{14!}{3!11!} = \dfrac{14\cdot 13 \cdot 12}{3\cdot 2} = 14\cdot 13 \cdot 2 = 364$