# Problem of the Week Problem D and Solution Number Display

## Problem

Helena’s Hardware Store is clearing out a particular style of single digits that are used for house numbers. There are currently only five $$5$$s, four $$4$$s, three $$3$$s, and two $$2$$s left.

How many different three-digit house numbers can be made using these single digits?

## Solution

Solution 1

Let’s suppose that there were three or more $$2$$s available. For the first digit, the customer could choose from the digits $$5,~ 4,~ 3,$$ and $$2$$. Therefore, there would be $$4$$ choices for the first digit. Similarly, there would be $$4$$ choices for the second digit, and $$4$$ choices for the third digit. This would give $$4\times 4 \times 4 = 64$$ possible three-digit house numbers that could be made.

However, there are actually only two $$2$$s available, so not all of these house numbers can be made. In particular, the house number $$222$$ cannot be made, but all others can.

Therefore, $$64 - 1 = 63$$ different three-digit house numbers can be made using these single digits.

Solution 2

Let’s look at three different cases.

Case 1: All three digits in the house number are the same
The house number could then be $$555$$, $$444$$, or $$333$$. The number $$222$$ cannot be made since only two $$2$$s are available. Therefore, there are $$3$$ three-digit house numbers with all three digits the same.

Case 2: Two digits are the same and the third digit is different
There are $$4$$ choices for the digits that are the same, namely $$5,~ 4,~ 3,$$ and $$2$$. For each of these possible choices, there are $$3$$ choices for the third different digit. For example, if two of the digits are $$5$$, then the third digit could be $$4,~ 3,$$ or $$2$$. Therefore, there are $$4\times 3 = 12$$ ways to choose the digits. For each of these choices, there are $$3$$ ways to arrange the digits. For example, suppose the digits are $$a,~a,$$ and $$b$$. The house number could be $$aab,~aba,$$ or $$baa$$.
Therefore, there are $$12\times 3 = 36$$ three-digit house numbers with two digits the same and one different.

Case 3: All three digits are different
The customer has $$4$$ choices for the first digit, namely $$5,~ 4,~ 3,$$ or $$2$$. Once that digit is chosen, there are $$3$$ choices for the second digit. Once the first and second digits are chosen, there are $$2$$ choices for the third digit. Therefore, there are $$4\times 3\times 2 = 24$$ three-digit house numbers with all three digits different.

Therefore, $$3 + 36 + 24 = 63$$ different three-digit house numbers can be made using these single digits.