[Scilab-users] Combinatorics

[Mikhail Urusov]

Dear Claus,

Just in case you have not got an answer yet:

> 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

This is right. In general, the formula for the number of k-combinations of
n elements is
n! / ( k! (n-k)! ).
Alternatively, you can write this as
n*(n-1)*....*(n-k+1) / (k!).
(This is the way you have written it above for n = 10 and k = 3.)
The explanation in general is quite similar to this case:

> 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?
For example, for n=10 and k=4:
If you consider "arrangements", i.e., the combinations, where the order
matters (e.g., 1-2-5-7 and 2-7-5-1 are different), then you have 10*9*8*7
such arrangements (for the first button you have 10 variants, for the
second one 9 variants, etc), but each "combination" (that is, where the
order does not matter) is counted 4! (=1*2*3*4) times (number of
permutations of 4 elements), so you need to divide: (10*9*8*7)/(4!).
For more information, see https://en.wikipedia.org/wiki/Combination

Best regards,
Mikhail


Am Do., 14. Apr. 2022 um 13:51 Uhr schrieb Claus Futtrup <cfuttrup at gmail.com
>:

> 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?
>
> 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
> _______________________________________________
> users mailing list
> users at lists.scilab.org
> http://lists.scilab.org/mailman/listinfo/users
>
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