SRL Archive - Help and Support

(This isn't homework)

Ok, if I have a plug board with 26 connections, and I have 13 connectors, how many possible connections can I make? All of the plugs need not be used.

Now, I have three wheels, with 26 possible positions. They are L(eft), M(idle), R(ight). When L goes past 26, M will move. When M goes past 26, R will move. How many combinations?

I have a separate stationary wheel, and it has 26 connections. it has 13 connectors… Except this time, each one must be used.

Serious question: Why would you possibly need to know this? Lol.

Sounds like homework to me.

1st one I think is ez:
26*13 = 338
as any combination can be used

2nd one isn't clear to me…
I can only see 26^3 = 17,576
I don't see how going pass 26 moves the other wheels as you said there is only 26 positions, 1-26 I'm guessing. So how would you go past position 26?

3rd one
(26*13)-13^2 = 169
I think thats how its done… Haven't tackled a problem like that that I recall.

The German Enigma machine had a plugboard with 26 connections, 13 connectors, three wheels, with each letter, and a reverser, that sent the message back through the wheels.



EDIT: Going past 26 resets the wheel at position one.

From: R0b0t1
(This isn't homework)

Ok, if I have a plug board with 26 connections, and I have 13 connectors, how many possible connections can I make? All of the plugs need not be used.

Now, I have three wheels, with 26 possible positions. They are L(eft), M(idle), R(ight). When L goes past 26, M will move. When M goes past 26, R will move. How many combinations?

I have a separate stationary wheel, and it has 26 connections. it has 13 connectors… Except this time, each one must be used.

1) = 10400600 possibilities? 26 nCr 13.
2)Not clear to me. I would say, 26^3.. Or, if you can only move 1 wheel at a time.. This would be, 26 * 3… When you pass 26, you start at 1 again, so it will be like this: 1..26, 1, 1 – 1, 1..26, 1 – 1, 1 , 1..26

3) Donno where ya talking about :rolleyes:

If I have a plugboard with 26 connections, and 13 connectors, how many combinations are left after all combinations that do not use all 13 connectors are eliminated?

And are you sure your right about #1?


EDIT: 2) If I have three wheels, each with 26 positions, how many combinations are there?

From: mastaraymond
1) = 10400600 possibilities? 26 nCr 13.
2)Not clear to me. I would say, 26^3.. Or, if you can only move 1 wheel at a time.. This would be, 26 * 3… When you pass 26, you start at 1 again, so it will be like this: 1..26, 1, 1 – 1, 1..26, 1 – 1, 1 , 1..26

3) Donno where ya talking about :rolleyes:

Ah right thats sounds better for #1
After i thought about it, it would have to be that high.