Evaluating n Terms in Equations with Addition

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In evaluating n terms in an equation using only addition, the number of valid ways to group the terms is determined by the arrangement of parentheses. For example, with four terms (a+b+c+d), there are five distinct ways to evaluate the expression by varying the grouping. The logic involves considering the order of operations and how parentheses can be placed around the terms. Additionally, the evaluation can differ based on whether the terms are kept in the same order or rearranged. Ultimately, the number of valid evaluations increases with the number of terms, highlighting the combinatorial nature of the problem.
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Given a number of terms, n in an equation containing only addition as the only possible operator, find the different number of valid ways in which they can be evaluated. Order of evaluation is controlled by grouping the terms in brackets

e.g. if n = 4
it means that there are 4 terms in the equation – i.e. something like
a+b+c+d

Now the valid ways in which it can be evaluated :
(a + (b + (c+d)))
(a + ((b+c) + d))
((a+b) + (c+d))
(((a+b) + c) + d)
((a + (b+c)) + d)

So the answer in this case is 5

what`s the logic to add n numbers in different possible ways?
 
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Two possible answers, depending on whether you keep them in the same order or not...

(a+b)+c, a+(b+c) two ways

(a+b)+c, (a+c)+b, (b+c)+a three ways
 
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