+207 Find the number of positive integers that satisfy both the following conditions:
Each digit is a $1$ or a $2$ or a $3$
The sum of the digits is $5$
+1908 We can use constructive counting to solve this problem.
CASE 1
First, let's see all the two digit numbers that satisfy these conditions.
We only have two, with 23 and 32.
CASE 2
Now, let's check all the 3 digit numbers.
We have 311, 131, 113, giving us 3.
We also have 221, 212, and 122, giving us another 3.
Case 3
4 digit number time! This is where it gets a bit complicated, but it's still not that bad.
We have numbers with three 1s and one 2, which has 4 different possibilities.
Case 4
Last case ! 5 digit numbers only have 1, with 11111.
Now, we just have to add up all our cases, and we get 2+3+3+4+1 = 13.
We have 13 numbers that satisfy the conditions.
NOTICE: This problem didn't have a lot of cases, so it would be easy to count. Bigger and wider ranges would require permutations!
Thanks! :)
