Le 14/04/2022 à 13:12, Claus Futtrup a écrit :
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
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?
--> nchoosek(10, 5)
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.
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:
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:
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
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.
users mailing list
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
users mailing list