$1+2+3+45+6+78+9=144$ what are other combinations Note that $$1+2+3+45+6+78+9 = 144$$ In how many other ways is it possible to make a total of $144$ using only $1, 2, 3, 4, 5, 6, 7, 8,$ and $9$ in that order and addition signs?
Sorry I am only in high school so dont over complicate the explanation. Thank you 

$123+4+5+6+7+8+9=162>144$.  Since any other way of having a three-digit number used in the summation will be strictly larger, we know none using a three-digit number exist.  Similarly, we know no arrangement with a number greater than three digits will yield a sum of $144$.
Using only one two-digit number, the largest it could be is $1+2+3+4+5+6+7+89 = 117<144$, so no arrangements with only one two-digit number exist.  Similarly, this also implies that no arrangements with all one-digit numbers exist since it will be smaller than even this.
Using two two-digit numbers, we already know that $1+2+3+45+6+78+9=144$ is a solution.
By changing it from $78+9$ to $7+89$ we have effectively increased the sum by $9$, similarly moving where the two-digit number is to the left will decrease the total by $9$, implying that $1+2+34+5+6+7+89=144$ is also a solution.
One can reason that any other pair of locations for the two-digit numbers will be either larger than or less than $144$
Using three two-digit numbers: through trial and error, we search for the one using the smallest available: $12+34+5+6+78+9=144$.
Using the same observation as before, we can shift one of the two digit numbers up and the other down to balance eachother out:  $12+3+45+67+8+9=144$
One can reason that no others exist using three two-digit numbers.
We look for four two-digit numbers: the smallest possible is $12+34+56+78+9=189>144$ so no more exist.
The complete list then:
$$1+2+3+45+6+78+9\\
1+2+34+5+6+7+89\\
12+34+5+6+78+9\\
12+3+45+67+8+9$$
for a total of four arrangements
