<s>[INST] $1+2+3+45+6+78+9=144$ 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 
 [/INST] 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  
</s>