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  =



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


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


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,


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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
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