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.