parenthesis of expression in such a way value not changed one example:

How many ways we can do possible value-preserving parenthesis the following expression in such a way that value not changed after parenthesis with one constraint that parenthesis among one variable is not correct (i.e: $(i)+j+k$ is not acceptable) ?

as an example for expression $i+j+k$, parenthesis like $i+(j+k)$, $((i+j)+k)$ are correct parenthesis but $(i)+j+k$ is not. I take a complete short example as follows.
for example if we have expression: $i+j-k$ there are $6$ possible (valid = value respect to original expression not changed) way that keep two condition given in question:
$i+j-k$
$(i+j)-k$
$(i+j-k)$
$((i+j)-k)$
$i+(j-k)$
$(i+(j-k))$
The answer for this question is not general and related to given expression.
Main Question:
expression: $ g-h*i+j-k/l/m$
Answer (according to my note): $96$

my challenge is how we get the point for the result? i.e: how this number calculated?

 A: I'm certain that the answer is $96$, but I'm not certain that the argument I will give will be considered fully complete.  First, let's look at the expression tree for the formula:

I apologize for the quality of the drawing; I don't have software to produce a neat picture easily, and I'm hopeless at drawing.  I have numbered the operations for later reference.  We can ignore them for now.
There are $6$ operations in the and we can parenthesize any of them that we like.  This gives $2^6=64$ possible parenthesizations.  Clearly all of them give equivalent expressions, because they don't change the expression tree.  When we parenthesize a particular operation, we say, in effect, "Consider the subtree rooted at this operation a unit."  But that subtree already is a considered as a unit.  The parentheses change nothing.
Please note that this considers the case of no parentheses as a valid answer, in contrast to one of your comments, but not to your little example.
Any parenthesization that resulted in an inequivalent formula would have a different expression tree, so if there were only one possible tree for the expression we would be done.  There is, however one other tree:

I've kept the labels on the operations the same as in the first tree, for ease of comparison.
We want to know if there are any ways to produce a parenthesization from the second tree that we didn't already produce from the first.  Comparing the two trees, we see that the subtrees rooted at operations, $2,\  4,\ 5$ and $6$ are identical in both trees, so any disparity must come from operation $1$ or $3$.  Parenthesizing operation $3$ in the second tree just introduces a a pair of parentheses around the whole expression, so it can't introduce a disparity that wasn't already there, and we are left with operation $1$.
Here we finally have a disparity.  Parenthesizing operation $1$ in the second tree gives the subexpression $$(g-h*i+j)$$ and there is clearly no way of producing this subexpression in tree $1$ because there is no subtree whose leaves are exactly $g,\ h,\ i$ and $j$.  Thus, and new parenthesization must parenthesize operation $1$.  We can parenthesize the other $5$ operations or not, as we will.  This gives the additional $2^5=32$ other parenthesizations.
The part of this argument that you, or your teacher may find incomplete is the assertion that there are only two possible expression trees.  This seems obvious to me, but you can supply a proof, if you want.  I would start by arguing that only operations $1$ and $3$ are possible at the root, and then for each case, argue that the rest of the tree is completely determined once the root is.
EDIT
Here is a proof that the two trees shown are the only possible ones.
Before we can do the last subtraction, we must have computed $k/l/m$, so neither operation $5$ nor $6$ can be at the root.  Before we can do first subtraction, we must have computed $h*i$, so operation $4$ cannot be at the root.  If operation $2$ were at the root, we would have $$g-[h*i+j-k/l/m),$$ which isn't equivalent because the signs change on the last two terms.  Thus, only operation $1$ or $3$ can be at the root.
If operation $1$ is at the root, we have $$[g-h*i]+[j-k/l/m]$$  On the left subtree, we've already seen that operation $2$ must come after operation $4$. On the right subtree, we've seen that operation $3$ must be the last one.  By convention, division associates from the left, so operation $5$ must come after operation $6$, and the right subtree is determined.
If operation $3$ is at the root, we have $$[g-h*i+j]-[k/l/m]$$  Since the order of operation in $k/l/m$ is fixed, the right subtree is determined.  On the left subtree, we know that operation $4$ cannot be at the root.  If operation $2$ were at the root, we would have $$[g]-[h*i+j],$$ and the sign on $j$ would change, so operation $1$ is at the root and the left subtree is $$[g-h*i]+[j]$$.  The right subtree is just a leaf.  On the left subtree, as we have seen before, operation $2$ must be the root.  These remarks show that the tree is determined.
In summary, the two pictured trees are the only expression trees possible.
