<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] 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.
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