TrojanPoem
  • TrojanPoem
Number of ways to multiply n matrices formula
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Michele_Laino
  • Michele_Laino
do we multiply \(n\) matrices?
TrojanPoem
  • TrojanPoem
The question asks for the number of ways to multiply the n matrices. for e.x: ABC = (AB)C = A(BC) (Number of ways = 2)
ganeshie8
  • ganeshie8
https://en.wikipedia.org/wiki/Catalan_number

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TrojanPoem
  • TrojanPoem
Thanks.
TrojanPoem
  • TrojanPoem
but I got a question, can I use this formula ? Combination of (number of choices) choose (two) = result result - 1
ganeshie8
  • ganeshie8
I feel the purpose of the question is to derive that formula, not simply to use it...
TrojanPoem
  • TrojanPoem
If so, I won't be able to prove your formula ( I won't copy & paste proof of course). well, how about this ? how to prove it ? https://allaboutalgorithms.files.wordpress.com/2011/11/parenthesis.jpg?w=595
Michele_Laino
  • Michele_Laino
Using a reasoning from statistical mechanics, I got this result: \[\Large \frac{{\left( {2n - 2} \right)!}}{{n!\left( {n - 1} \right)!}}\] it works for \(n=1,2\)
Michele_Laino
  • Michele_Laino
where, of course, \(n\) is the number of the involved matrices
TrojanPoem
  • TrojanPoem
n = 3 , result = 2 n = 4, result = 5 Check your formula.
Michele_Laino
  • Michele_Laino
for \(n=4\), we get: \[\Large \frac{{\left( {2n - 2} \right)!}}{{n!\left( {n - 1} \right)!}} = \frac{{\left( {8 - 2} \right)!}}{{4! \cdot 3!}} = \frac{{6!}}{{4! \cdot 6}} = \frac{{4! \cdot 5 \cdot 6}}{{4! \cdot 6}} = 5\]
Michele_Laino
  • Michele_Laino
whereas for \(n=2\), we get: \[\Large \frac{{\left( {2n - 2} \right)!}}{{n!\left( {n - 1} \right)!}} = \frac{{\left( {4 - 2} \right)!}}{{2! \cdot 1!}} = \frac{{2!}}{{2 \cdot 1}} = \frac{2}{2} = 1\]
Michele_Laino
  • Michele_Laino
finally, for \(n=3\), we can write: \[\Large \frac{{\left( {2n - 2} \right)!}}{{n!\left( {n - 1} \right)!}} = \frac{{\left( {6 - 2} \right)!}}{{3! \cdot 2!}} = \frac{{4!}}{{12}} = \frac{{24}}{{12}} = 2\]
TrojanPoem
  • TrojanPoem
Awesome! nice indeed.
TrojanPoem
  • TrojanPoem
Is there a way to verify if it will lag in bigger numbers ?
TrojanPoem
  • TrojanPoem
How about verifying it using this formula ? https://allaboutalgorithms.files.wordpress.com/2011/11/parenthesis.jpg?w=595
Michele_Laino
  • Michele_Laino
I don't know. I have used the same reasoning in order to get the \(Bose-Einstein\) statistics
TrojanPoem
  • TrojanPoem
Bose - Einstein ? what is that ?
Michele_Laino
  • Michele_Laino
it is a topic of Statistical Mechanics!
TrojanPoem
  • TrojanPoem
So it's right :),
Michele_Laino
  • Michele_Laino
if we make this variable change: \(m=n-1\), we get: \[\huge \frac{{\left( {2n - 2} \right)!}}{{n!\left( {n - 1} \right)!}} = \frac{{\left( {2m} \right)!}}{{\left( {m + 1} \right)!m!}}\] namely, the same formula of @ganeshie8
TrojanPoem
  • TrojanPoem
It's better for on the fly questions , Thanks a lot.
Michele_Laino
  • Michele_Laino
:)

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