$1+2+3+45+6+78+9=144$ 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 
 A: $123+4+5+6+7+8+9=162>144$.  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
A: If we "bunch up"  $8$ and $9$ to get $89$, the rest of the numbers must add up to $55$.  Bunching $6$ and $7, 5$ and $6$, and $4$ and $5$ will make the sum too high.  If we bunch $3$ and $4$ to get $34$, so we have $34+5+6+7+89=141$ which needs $3$ more.
So $$1+2+34+5+6+7+89=144$$.
That's the only possible $89$ bunch up as $12+3+4+5+6+7+89$ doesn't work.
You have a solution with a $78$ bunch up. Any other $78$ bunch up has $78+9=88$ requires the rest of the numbers to add to $56$.  We can't bunch $56$.  If we bunch up $45$ we get $45+6+78+9=138$ and we need $6$ more so we get your answer if we don't bunch up $45$ and bunch up $34$ we have $34+5+6+78+9=132$ and we need $12$ so we can do $12+34+5+6+78+9$.
If we bunch up $67$ we get $67+8+9=84$ and we need $60$ more.  If we bunch $45$ we need $15$ more from $1,2,3$ so we can have $12+3+45+67+8+9$.
If we bunch $67$ but not $45$ we have $5+67+8+9=89$ so we need to get $51$ from $1,2,3,4$.  The most we can do is $12+34$ so there is no other with $67$ bunch.
If we bunch up $56$ we have $56+7+8+9=80$ and we need $64$ more.  If we bunch up the $34$ we need $30$ more which can't be done with only the $1$ and $2$.
If we don't bunch the $56$ we have $5+6+7+8+9=35$ and we need to get $109$ from $1,2,3,4$.  The most we can get is $12+34=46$ so we can't go any further.  Those four are the only solutions.
A: Inspired by the idea: "1+2+..9=45 therefore we need 99 more to get 144. The two digits 'ab'=10a+b so changing a+b to 'ab' adds 9a to the total. Therefore an extra eleven 9s are required. This means that the the possibilites are (3,8) (6,4,1) etc."
We need find all the combinations that add up to 11. But in these combinations, the difference of any two numbers must >=2. Otherwise, we can not make these two 2-digit numbers at the same time, for example, 5+6=11, but we can not make numbers 56 and 67 at the same time, as 6 has been used twice. Also we can not use number 9, as after 9, there is no more number to make 2-digit number 9x.
The following are all the possible combinations:
4+7=11
3+8=11
1+3+7=11
1+4+6=11
Therefore all the correspondent answers are:
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  
