[Scilab-users] Combinatorics

[Stéphane Mottelet]

Hi,

Le 14/04/2022 à 13:12, Claus Futtrup a écrit :
>
> Dear Scilabers
>
> I hope you can help me out. My combinatorics is a bit rusty.
>
> So, the spouse has purchased a lock and I wondered how many 
> combinations are available?
>
> The lock has 10 push buttons, they are numbered 1-2-3-4-5-6-7-8-9-0.
>
> From a programming point of view, any of the numbers can be set on or 
> off, meaning there are 2^10 = 1024 combinations, as far as I can see.
>
> I wonder how they are distributed, and how many of the numbers I 
> should activate in the lock to maximize the number of combinations?
>
The number of different subsets of k distincts elements of a set 
composed of n elements is the binomial coefficient (n,k). When n is 
even, it is maximized for k=n/2. Here

-->  nchoosek(10, 5)
  ans  =

    252.

S.

> Let's see, we have:
>
> None (none of the buttons are activated), there's exactly 1 
> combination for this situation. The lock is delivered from the 
> manufacturer in this state.
>
> All (all of the buttons are activated), there's exactly 1 combination 
> for this situation as well (no variability).
>
> One button pushed. There's obviously 10 possible combinations (push 
> any one of the 10 buttons).
>
> Two buttons pushed. There's 10 * 9 / 2 = 45 combinations. Each button 
> can only be pushed once, so once you've selected the first button, 
> there's only 9 left, but also we divide by two because the combination 
> are doubled, I mean for example the combination 1-2 = 2-1 ... the lock 
> doesn't know the difference. If you spread out the possibilities in a 
> 2D plane, it's like ignoring the diagonal (like pushing the same 
> button twice) and also we either ignore the upper or lower triangle. 
> Makes sense?
>
> Here starts my trouble. Three buttons pushed. Instead of looking at a 
> 2D plane, I guess you spread out in 3D. The diagonal line is more than 
> that - we have several planes where two of the three numbers are the 
> same (and which are not allowed).
>
> To help myself out, I've tried to write all combinations where one of 
> the push buttons is number 1. We select all combinations with the 
> second button being either 2-3-4 and so on, and how many combinations 
> do we then have for the third option? See table below:
>
> 1-2-x 	8
> 1-3-x 	7
> 1-4-x 	6
> 1-5-x 	5
> 1-6-x 	4
> 1-7-x 	3
> 1-8-x 	2
> 1-9-0 	1
>
> 	36
>
> We can then do the same for the first button = number 2, and we get : 
> 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 combinations and so on. We get:
>
> 1-x-y 	36
> 2-x-y 	28
> 3-x-y 	21
> 4-x-y 	15
> 5-x-y 	10
> 6-x-y 	6
> 7-x-y 	3
> 8-9-0 	1
>
> 	120
>
> OK, so that was with three buttons pushed. It's always good to know 
> the answer (if it's correct :-/ I hope it is), but it's a tedious 
> process and I was wondering if you could point me to an easy 
> calculation instead? ... Ideally something that expands to 4 and 5 
> buttons.
>
> I can 'invent' a calculation, which could be : =10*9*8/(3*2*1) = 120 
> ... if this indeed shows the internal workings, I'd like to know why. 
> Sorry my combinatorics is so bad ... I haven't played in this field 
> for a while.
>
> Best regards,
>
> Claus
>
>
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-- 
Stéphane Mottelet
Ingénieur de recherche
EA 4297 Transformations Intégrées de la Matière Renouvelable
Département Génie des Procédés Industriels
Sorbonne Universités - Université de Technologie de Compiègne
CS 60319, 60203 Compiègne cedex
Tel : +33(0)344234688
http://www.utc.fr/~mottelet
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