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Question: Write out all of the multiplication schemes for four numbers $a_1,a_2,a_3,a_4$ in that order.

My Attempt: By explicating listing the elements we get the answer as $5$.

$$ (((a_1a_2)a_3)a_4) $$ $$ ((a_1a_2)(a_3a_4)) $$ $$ ((a_1(a_2a_3))a_4)$$ $$ (a_1((a_2a_3)a_4))$$ $$ (a_1(a_2(a_3a_4))) $$

Doubt: Is my answer correct? Is there a better way of doing this?

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  • $\begingroup$ Tried permutation via gap method? $\endgroup$ – The Dead Legend Apr 11 '17 at 0:37
  • $\begingroup$ AND your 1st and 3rd combination are the same. $\endgroup$ – The Dead Legend Apr 11 '17 at 0:37
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    $\begingroup$ There is a better way as you suspect. You are looking for Catalan numbers $$C_n=\frac{1}{n+1}\binom{2n}{n}$$ in this case there are $3$ pairs of brackets so $n=3$ and $C_3=\frac{1}{4}\binom{6}{3}=5$ as you correctly state. $\endgroup$ – N. Shales Apr 11 '17 at 1:37
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    $\begingroup$ @N.Shales Thanks man. This was what I was looking for. $\endgroup$ – user330477 Apr 11 '17 at 2:01
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    $\begingroup$ Your answer, 5, is correct. Beyond brute force listing them all you might want to check the Catalan numbers: en.wikipedia.org/wiki/… $\endgroup$ – Ethan Bolker Apr 11 '17 at 12:24
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I don't know how much will this help, but consider this
$$(X,a_{1},X,a_{2},X,a_{3},X,a_{4},X)$$ suppose we have to put 2 braces . We will take any 2 positions out of 5. Using 5C2. Now suppose you put 2 around $a_{1}$ and $a_{3}$. It will look like, $$(X,a_{1},X,a_2,X,a_3,X)a_4$$

Now again the same situation, 2 braces. 4 places. : 4C2
One may go on and on and on..(One may improvise on this one. Cheers)

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  • $\begingroup$ This is not my question mate. An opening bracket also has to end $\endgroup$ – user330477 Apr 11 '17 at 0:57
  • $\begingroup$ Yes, that is what I am talking about? 2 braces: '(' and ')' $\endgroup$ – The Dead Legend Apr 11 '17 at 1:01
  • $\begingroup$ Does this method ultimately give $5$, which is indeed correct? $\endgroup$ – pjs36 Apr 11 '17 at 1:37

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