Originally Posted by

**Soroban** Hello, Ragnarok!

I have a broader interpreation of this problem.

If I'm right, the problem is quite complex.

$\displaystyle \begin{array}{cccc}\text{Divide the factors into 5 groups:} &

(a)(b)(c)(d)(e) \end{array}\quad \text{1 way}$

$\displaystyle \text{Divide the factors into 4 groups: }\;\begin{array}{c}

(ab)(c)(d)(e) \\(a)(bc)(d)(e) \\ (a)(b)(cd)(e) \\ (a)(b)(c)(de) \end{array}\quad \text{ 4 ways}$

$\displaystyle \text{Divide the factors into 3 groups: }\;\begin{array}{c}

(abc)(d)(e) \\ (ab)(cd)(e) \\ (ab)(c)(de) \\ (a)(bcd)(e) \\ (a)(bc)(de) \\ (a)(b)(cde) \end{array}\quad \text{ 6 ways}$

$\displaystyle \text{Divide the factors into 2 groups: }\;\begin{array}{c}

(a)(bcde) \\ (ab)(cde) \\ (abc)(de) \\ (abcd)(e) \end{array} \quad \text{ 4 ways}$

$\displaystyle \text{Divide the factors into 1 group: }\;\;\;\begin{array}{c}(abcde)\end{array} \quad \text{ 1 way}$

. . There are: .$\displaystyle 1 + 4 + 6 + 4 + 1 \:=\:16$ ways.

$\displaystyle \text}For }\,n\text{ factors, there are }2^{n-1}\text{ ways to insert parentheses.}$

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However, *my* interpretation opens the door to another realm,

. . one I do *not* want to explore . . .

Are we to consider groupings like these?

. . $\displaystyle \bigg((ab)(c)\bigg)(d)(e) $

. . $\displaystyle (ab)\bigg((c)(d)\bigg)(e) $

. . $\displaystyle (ab)(c)\bigg((d)(e)\bigg) $

I hope not . . . *Brrr* . . .