Flowerfred

Not that worried to be honest, as the two most popular ways to steal a car are still:

1 breaking into the house for the keys

2 picking up the entire car, either on a flatbed or with specially prepared trailers hat just fit over the car and lift it off the ground.

This seems way too high tech for the average junky that's after a few cd's or a set of golf clubs.

Rudy

The physical cut of the key is the same beteween the door and the starter but there is an electronic chip inside the key that is needed before the car can start. All this tool does is allow you to turn the lock cylinder (which would allow access to the car). If you tried to use it to start the car, it wouldn't work.

Then again, there's probably a tool out there to duplicate the electronic chip...

530i_msport

I was talking about the cars with the button.

Previous models should be accessible as the key is the same.

530i_msport

ok Rudy was faster

BangleBox_530d

AFAIK the "push button" models are just as insecure. A bit more hi-tech to steal the key code, but far from secure. I heard that the Beckhams lost both X5s from the key being decoded.

aybeesea

That indeed would be the answer if each item could be chosen more than once.

If you mean permutations (where order matters and no item can be chosen more than once) then permuting 2 from 16 gives 240 permutations.

If you mean combinations where order doesn't matter then combining 2 from 16 gives 120 combinations.

Actually there are 32 individual half keys so the calculation is "2 from 32".

Permutations: 992

Combinations: 496.

Since the order of left and right is significant I make 2 from 32 choices 992 permutations.

Or I may have misunderstood of course.

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BetterMakeWay

WOW indeed impressing, but one has to have the chip from inside the key to actually start the car right?! Either way it's indeed scary of how easy it can be done...I'm wondering now if the actual start/stop pushing button is indeed safer than some of the old fashion laser cut key system.

OK later edit: I notice Rudy already answered it. Then it appears both systems have their flaws but which one in the end would be safer?! I mean how many thiefs run around with super high tech equipment to steal codes and that stuff...?!

dlevi67

I'm afraid you have. Each half key code (starting from 1111 ending with 2222) can be either left or right, including cases where you have the "same" half key on both left and right. This gives exactly 240 permutations + 16 cases where the same "sequence" is picked twice, once for left, once for right. There aren't 32 equivalent half keys, because you cannot use the "right half" key in the left half of the lock (amongst other things, it won't even enter it, as there are the guiding tags in the way!).

If you prefer, think of it another way: this is equivalent to counting how many numbers you can make in hex with two digits 00 to FF (if you re-number the '1' to 0 and the '2' to 1, or viceversa, you get the two digits in binary). Surprise surprise, the answer is FF+1 = 256. (Or, for those of us that do not do hex operations as quickly as decimal, how many numbers can you compose using two digits from the 0-9 set. The answer is 100, not 90, which is the number of permutations that you get as 10! / (10-2)! )

aybeesea

Hmm... I don't understand your first paragraph explanation.

If as you say left and right can only be selected for such sides of the key then that reduces the choices dramatically. i.e 16 choices per side.

Of course that then gives 16^2 choices when you combine the two choices for both sides and bingo 256! (edit: I notice that you said (4^2)^2) earlier which seems a complex way of expressing a simple idea)... But I really DON't understand "This gives exactly 240 permutations + 16 cases where the same "sequence" is picked twice, once for left, once for right."

So let's agree on 256 but not on the logic or maths that gets us there.

*Delete*

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aybeesea

For those of us that don't understand permutations, you cannot use the same symbol (or physical item) AGAIN once it has been chosen

So, 00, 11, 22 ... 99 are EXCLUDED in the calculation of permutations. Ir's not surprising that there are 90 permutations when these occurrences are removed.

I'm not sure where using permutations in the explanation of number bases is taking us.

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