Find all ways of grouping five multiplied numbers

I am working through Israel Gelfand's book "Algebra" and am stuck on the following problem:

*Find all possible ways to put parentheses in the product 2*3*4*5*6 (not changing the order of factors). Try to invent a systematic way of searching so as not to forget any possibilities.*

I calculated 16 possibilities by rote before giving up (there may be more), and the only thing I accomplished in the way of "systematic" was stating that in any sequence of five numbers there are four operations to be performed. Perhaps it's the late hour, but I can't get beyond this. Any suggestions?

Thanks!

Re: Find all ways of grouping five multiplied numbers

Quote:

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.

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

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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* . . .

I think you are correct in this one, I am also kind of stuck with this problem in Gelfand's book and I think I counted every possibility for 2*3*4*5*6 and I always get 16 an result which can also be calculated by 2^{n-1}. So thank you for helping.