License plates in Riddleland

anna_gg · 2016-01-14 10:27:01

In Riddleland, the license plates are composed of 8 digits, each from 0 to 9. All plates must differ from each other in at least 2 of the 8 positions. What is the maximum number of plates that can be issued?

phrontister · 2016-01-14 16:24:51

Hi anna_gg,

The wording seems to allow a plate to have repeat digits (eg, 12345677). Is that so, or must the 8 digits differ from each other (eg, 12345678)?

Last edited by phrontister (2016-01-14 22:20:20)

anna_gg · 2016-01-15 10:16:48

Repetition of digits is allowed.

phrontister · 2016-01-17 00:53:03

Hi anna_gg,

Do you know the answer if there is no digit repetition?

anna_gg · 2016-01-18 11:35:34

Unfortunately not!

phrontister · 2016-01-18 12:56:19

Oh...pity.

I'd been working like this, using the 'no repetitions' and 'with repetitions' formulas from MIF's page here:

In both formulas, "n is the number of things to choose from, and we choose r of them".

So, simply, n = 10 (the number of digits in 0-9) and r = 8 ("the license plates are composed of 8 digits").

However, the constraint that "all plates must differ from each other in at least 2 of the 8 positions" has to be included in the formula somehow, and I thought that r would be the place to do it.

I reasoned r this way:
- I think the maximum number is obtained with the least number of differences, and so I ignored looking for numbers of positions greater than the least (2).
- If all plates differ from each other in 1 of the 8 positions, then r = 8.
∴ if all plates differ from each other in 2 of the 8 positions, then r = 8-1 = 7.

No repetitions:

With repetitions:

Maybe my r is wrong. I have the feeling that the reasoning for r should be more complex than simply deducting 1 from the choice of 8 digits.

Don't know enough permutations theory, though, and I haven't been able to work out where I went wrong...

Last edited by phrontister (2016-01-20 02:29:52)

anna_gg · 2016-01-22 21:18:22

@Phrontister: It seems your result is correct, but through some other solution
For the first 7 digits, obviously the maximum number of possible different plates with at least one different digit in one position (given that each digit can have any value from 0 to 9 and repetition IS permitted), is 10^7.
Now we must find a way to add an 8th digit in such a way that each plate is different from each other also in the last digit. That is, if we consider two (7-digit) plates that only differ in 1 of the 7 digits, the 8th must be selected in such a way that it always creates a different 8-digit plate. We are therefore looking for a checksum - for example calculating the sum of the first 7 digits --> MOD(10) and then selecting the last digit to be -MOD(10), so that the sum of all 8 digits is always a product of 10.
So in effect, yes, the maximum number is 10^7!

Last edited by anna_gg (2016-01-23 00:38:23)

phrontister · 2016-01-23 05:23:43

Hi anna_gg;

I like your method...nice and logical.

I ran yours in an Excel spreadsheet and it checks out perfectly. Couldn't go the whole 10,000,000 because of Excel's row limit of 2^20, but went to row 1,000,000 just for fun even though I got the picture miles earlier than that.

Mine is the simpler of the two methods, but it may just have been a fluke to have worked (although I did my best to reason it out). Maybe I'll think about it some more...

Math Is Fun Forum

#1 2016-01-14 10:27:01

License plates in Riddleland

#2 2016-01-14 16:24:51

Re: License plates in Riddleland

#3 2016-01-15 10:16:48

Re: License plates in Riddleland

#4 2016-01-17 00:53:03

Re: License plates in Riddleland

#5 2016-01-18 11:35:34

Re: License plates in Riddleland

#6 2016-01-18 12:56:19

Re: License plates in Riddleland

#7 2016-01-22 21:18:22

Re: License plates in Riddleland

#8 2016-01-23 05:23:43

Re: License plates in Riddleland

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