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 manycombinations 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 oroff, 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 Ishould 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 1combination for this situation. The lock is delivered from themanufacturer in this state.All (all of the buttons are activated), there's exactly 1 combinationfor this situation as well (no variability).One button pushed. There's obviously 10 possible combinations (pushany one of the 10 buttons).Two buttons pushed. There's 10 * 9 / 2 = 45 combinations. Each buttoncan 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 combinationare doubled, I mean for example the combination 1-2 = 2-1 ... the lockdoesn't know the difference. If you spread out the possibilities in a2D plane, it's like ignoring the diagonal (like pushing the samebutton twice) and also we either ignore the upper or lower triangle.Makes sense?Here starts my trouble. Three buttons pushed. Instead of looking at a2D plane, I guess you spread out in 3D. The diagonal line is more thanthat - we have several planes where two of the three numbers are thesame (and which are not allowed).To help myself out, I've tried to write all combinations where one ofthe push buttons is number 1. We select all combinations with thesecond button being either 2-3-4 and so on, and how many combinationsdo 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 36We 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 120OK, so that was with three buttons pushed. It's always good to knowthe answer (if it's correct :-/ I hope it is), but it's a tediousprocess and I was wondering if you could point me to an easycalculation instead? ... Ideally something that expands to 4 and 5buttons.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 fieldfor a while.Best regards, Claus _______________________________________________ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users

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