# Casework counting and overcounting: How many numbers can be obtained by adding certain partitions of a set

Using each of the digits $\{1, 2, 3, 4\}$ exactly once and using zero or more $+$ signs, how many different totals can be obtained? No digit can be used as a power, root, etc. Only addition can be done. When $n$ digits are written next to each other, they represent an $n$-digit number and not the product of $n$ factors. Four of the totals to be included are $\underline{1234}$, $\underline{55} = 12 + 43$, $\underline{127} = 123 + 4$ and $\underline{37} = 31 + 2 + 4$.

My reasoning: There are two cases we need to consider.

1: We can find numbers such as $1234$, $2341$, etc. We can split this into a few subcases. There are $4$ one-digit numbers that can be obtained by this (Note: I am not counting $5$, $6$, $7$...Although they can be obtained, they are not part of this case). Next, we can find two digit-occurrences. This would be quite simple. We can partition the set $\{1,2,3,4\}$ into two two-digit numbers. (e.g 12+34, 31 + 24).

We can list these numbers out. 12,13,14,21,23,24,31,32,34,37,41, 42, 43, 46,55,64,73 (I believe these are all of the cases. Correct me if I'm wrong). HEREIN LIES THE PROBLEM. We can also partition the set into one two-digit number and one (or two) one-digit numbers. This could be a problem.

Now, we can get to 3-digit numbers. There is only one way to obtain this. We partition the same set into one three-digit number and a one-digit number. This gives us $123+4$, $132+4$, and $142+3$ for the cases where $1$ is the hundreds digit. We can multiply this by 4, to get $12$ three-digit numbers. We could also take 3 numbers from the set and rearrange them. But, that would cause overcounting.

Next, we get to 4-digit numbers. This is simple. We just take the $4!$ ways there are to rearrange 1234. This gives us 24 + 17 + 12 = 53 cases. However, there are more.

Case 2: We can just start adding numbers...

I REALLY NEED HELP...It's appreciated.

There will be 0, 1, 2, or 3 plus signs. If there are none, there are $4!=24$ possible results; if there are three, there is only $1$ possible sum, ten. Now we consider the other cases.

For reference, there are only five possible results when adding two of these digits together. $$1+2=3\qquad 1+3=4\qquad 1+4=2+3=5\qquad 2+4=6\qquad 3+4=7$$

One plus sign. Suppose we have $ABC+D=ABE$. If $E=5$, then there are two possible values for the set $\{A,B\}$, otherwise only one. This gives $2+2+4+2+2=12$ different sums.

Suppose instead we have $AB+CD = EF$. Then $\{E,F\}$ is one of $\{3,7\},\{4,6\},\{5\}$. This gives $2+2+1=5$ different sums.

Two plus signs. We must have $AB+C+D = AE$. There are $4$ choices for $A$, but then $A+E=10$ so $E$ is determined, and there is no carry in the addition. Thus there are only $4$ different sums. Of these, $37$ and $46$ were already counted above.

Thus there are $24+12+5+4-2+1=44$ different possible sums. Only twenty of those are less than $1234$.