Find the number of positive integers that satisfy the following conditions:
* Each digit is a 1 or a 2 or a 3
* The sum of the digits is 5
+195 We can use casework to solve this problem. We need to find all the ways to arrange the digits 1, 2, and 3 to sum to 5.
Case 1: One digit is 5
There is only one way to arrange the digits in this case: 5. This is not a valid solution since the digits must be 1, 2, or 3.
Case 2: Two digits sum to 5
The two digits that sum to 5 can be either 2 and 3 or 3 and 2. There are two ways to arrange each pair of digits: 23 or 32. We need to determine how many times each pair appears.
For 23, we can choose any two of the five digits to be 2, and the remaining digit must be 1. There are 5 choose 2 = 10 ways to choose the positions of the 2's, and the remaining digit must be 1. Therefore, there are 10 numbers that have two digits that sum to 5 and contain the digits 1, 2, and 3: 221, 212, 122, 233, 323, 332, 131, 113, 311, and 212.
For 32, we can choose any two of the five digits to be 3, and the remaining digit must be 2. There are 5 choose 2 = 10 ways to choose the positions of the 3's, and the remaining digit must be 2. Therefore, there are 10 numbers that have two digits that sum to 5 and contain the digits 1, 2, and 3: 223, 232, 322, 233, 323, 332, 221, 212, 122, and 131.
Case 3: Three digits sum to 5
The three digits that sum to 5 can only be 1, 2, and 2. There is only one way to arrange these digits: 122. We need to determine how many times this arrangement appears.
We can choose any three of the five digits to be 2, and the remaining digits must be 1. There are 5 choose 3 = 10 ways to choose the positions of the 2's, and the remaining digits must be 1. Therefore, there are 10 numbers that have three digits that sum to 5 and contain the digits 1, 2, and 3: 122, 212, 221, 112, 121, 211, 131, 113, 311, and 313.
Therefore, there are a total of 10 + 10 + 1 = 21 numbers that satisfy the conditions.
