How many combination possibilities in homemade lock

[davidbenjamindix]

I am designing a two part magnetic lock for my blanket chest. It is a two-part lock. The first part has 3 master buttons (primary buttons), and the second part has 11 secondary buttons. What you do first is choose one of the master buttons (it stays depressed). Then out of the secondary buttons, you choose three that will open the lock. So, not only do you have to choose the correct master button, but the 3 secondary buttons as well. Each button is a spring-loaded buttons, so it pushes back out when you stop pressing it, and inside the lock each button has a magnet that controls the opening/closing of the lock. How many possibilities are there in this lock?

[davidbenjamindix]

is the answer 531,444? (3 to the eleventh + 3 to the eleventh + 3 to the eleventh)??

[zeepia]

I would say 3993 (3 x 11 x 11 x 11) but somebody who knows his math could tell the right answer...

How the magnets work in that system?

[davidbenjamindix]

30 neodymium rare earth N26 1/2" x 1/8" discs (super strong)

Depending on which button is pressed, the magnets either attract or repel towards or against each other. These types of locks are exceedingly difficult and tedious to make as you would imagine. It takes a ton of planning.

[davidbenjamindix]

[quote="zeepia"]I

Oops. I misunderstood your question. I thought you asked how many magnets, rather how do the magnets work. sorry.

This is how it works. The push buttons are horizontal, and are imbedded into the buttons (made out of wild cherry hardwood) by drilling 1/2" x 1/8" holes and superglued into place. For the lock mechanism, I also imbed a magnet. If it is a push button that is meant to lock, then I will use a repelling magnet (+ against +). If it is meant to attract (to open that part of the lock) I will use + against - , or - against +.

So when you push the button that is horizontal, it forces or attracts the lock component which is vertical and also has a magnet in it. Only three of these secondary blocks will repel blocks that open the lock, while the other 8 make sure the lock stays lock by pulling the lock components closed. The master controls (also three magnetic/spring mechanisms) are made in the same manner. One of those masters is an unlocker, while the other two lock. The master controls stay depressed with a level that you lock into place after depressing it. That's pretty much how it will work. All of this is only one part of the whole lock. Solving this puzzle gives you a magnetic key card which actually unlocks the blanket chest. There is also ANOTHER puzzle before . It is a two-part complicated word riddle. If you can solve both riddles, you will have both the answer to the master and secondary buttons.

[MacGyver101]

The first part has 3 master buttons (primary buttons), and the second part has 11 secondary buttons. What you do first is choose one of the master buttons (it stays depressed). Then out of the secondary buttons, you choose three that will open the lock. So, not only do you have to choose the correct master button, but the 3 secondary buttons as well.

Assuming that it doesn't matter what order you press the "secondary buttons" in, then you have a total of 495 possible combinations:

3 primary buttons x ( 11 secondary buttons choose 3 )
= 3 x ( 11! / (3! x (11-3)! )
= 3 x 165
= 495

...although I'm slightly overdue for my afternoon coffee.

[GWiens2001]

You would probably not want to have all of the opening or closing buttons either repelling or attracting, but rather a mix, same with the non-functional buttons. If the buttons that open/close all feel different than those that do not, it would be easy to tell which buttons were needed. That would only leave the order to be deciphered.

Just my 2 cents

Gordon

[maintenanceguy]

For the primary buttons,there are 3 options: button a, buttton b, or button c: So that portion gives 3.

For the eleven remaining buttons, the first button chosen has 11 possible options. The second button has 10 remaining possibilities, and the 3rd button has 9 remaining possibilities. So that part has 11x10x9 = 990 combinations.

3x990 = 2,970 possible combinations.

--------

This is only true if everybody knows you need to pick one of the first 3 followed by 3 of the remaining 11. And if you can not choose the same button twice.

If the number of buttons to choose is unknown, than you have to do the above math for the possibility of 1 button, 2 buttons, 3 buttons all the way up to all 11 buttons and multiply that by the options of choosing 1, 2, or 3 of the first buttons.

I get 759,535,777 combinations - assuming you can only use each button once.

[zeepia]

I love it when everybody has their own solution for that combination count

You must definitely show us pictures of your lock when it´s finished, it sounds inspiring!

[MBI]

42 possible combinations.

[MacGyver101]

For the eleven remaining buttons, the first button chosen has 11 possible options. The second button has 10 remaining possibilities, and the 3rd button has 9 remaining possibilities. So that part has 11x10x9 = 990 combinations.

3x990 = 2,970 possible combinations.

As you noted, that's the right answer if the order of pressing the second buttons matters (i.e., if pressing [A]-[B]-[C] is different than pressing [C]-[B]-[A] or [B]-[A]-[C], etc.). From the description of the mechanism, though, it doesn't sound like the order matters... so, if the combination is 1 "primary" button and 3 "secondary" buttons, then I'm pretty certain there are only 495 combinations.

If you're choosing any number of "primary" buttons (1, 2 or 3) and any number of secondary buttons (anywhere from 1 to 11), and the order of pressing them doesn't matter, then you have:

(3C1 + 3C2 + 3C3) x (11C1 + 11C2 + 11C3 + ... + 11C11)
= (3 + 3 + 1) x (11 + 55 + 165 + 330 + 462 + 462 + 330 + 165 + 55 + 11 + 1)
= 7 x 2047
= 14329 combinations

[minifhncc]

(3C1 + 3C2 + 3C3) x (11C1 + 11C2 + 11C3 + ... + 11C11)
= (3 + 3 + 1) x (11 + 55 + 165 + 330 + 462 + 462 + 330 + 165 + 55 + 11 + 1)]

Lol, you don't need to sum up manually like that...

Use the fact that (1+x)^n = nC0 + nC1 x + nC2 x^2 +... + nCn x^n

Sub x=1 and rearrange a bit : 2^n - 1 = nC1 + ... + nCn

[zeepia]

And same in english, please

[MacGyver101]

Lol, you don't need to sum up manually like that...

I know... but just saying "((2^11)-1)" wouldn't help someone understand how I came up with the number, or demonstrate how it relates to the lock that we're talking about.

[GWiens2001]

42 possible combinations.

But that answer only applies on Thursdays.

Gordon