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Science Forum Index » Statistics - Math Forum » Licence plate simple question
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| Francogrex |
 Posted: Wed Apr 16, 2008 1:53 am |
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Hi, please check if I'm doing this right: if we have a licence plate
of the type 3 letters-3 integers (ex: abc123) then the total number of
unique plates would be equal to 26^3*10^3= 17,576,000. Right?
But what if we allow the 3 letters and the three integers to be in any
order (ex:a12b3c or 1ab32c etc), would the total numbre be
(26^3*10^3)*6= 105,456,000?
Thanks. |
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| Paul Rubin |
| Posted: Wed Apr 16, 2008 7:32 am |
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Francogrex wrote:
Quote: Hi, please check if I'm doing this right: if we have a licence plate
of the type 3 letters-3 integers (ex: abc123) then the total number of
unique plates would be equal to 26^3*10^3= 17,576,000. Right?
Yes.
Quote: But what if we allow the 3 letters and the three integers to be in any
order (ex:a12b3c or 1ab32c etc), would the total numbre be
(26^3*10^3)*6= 105,456,000?
Thanks.
No. First, the number of permutations is 6!, not 6. Second, you have
to account for indistinguishable permutations (e.g., AAA111 has only 20
distinguishable permutations).
/Paul |
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| Stan Devia |
| Posted: Wed Apr 16, 2008 7:59 am |
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On 2008-04-16 21:23:59 +0930, Francogrex <franco@grex.org> said:
Quote: Hi, please check if I'm doing this right: if we have a licence plate
of the type 3 letters-3 integers (ex: abc123) then the total number of
unique plates would be equal to 26^3*10^3= 17,576,000. Right?
But what if we allow the 3 letters and the three integers to be in any
order (ex:a12b3c or 1ab32c etc), would the total numbre be
(26^3*10^3)*6= 105,456,000?
Thanks.
There are 6!/3!3! = 20 permutations of L L L N N N where L=letter and
N=number so the total number would be 26^3*10^3*20 = 351,520,000 |
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| Stan Devia |
| Posted: Wed Apr 16, 2008 4:11 pm |
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On Apr 17, 10:31 am, Richard Ulrich <Rich.Ulr...@comcast.net> wrote:
Quote: On Wed, 16 Apr 2008 04:53:59 -0700 (PDT), Francogrex <fra...@grex.org
wrote:
Hi, please check if I'm doing this right: if we have a licence plate
of the type 3 letters-3 integers (ex: abc123) then the total number of
unique plates would be equal to 26^3*10^3= 17,576,000. Right?
But what if we allow the 3 letters and the three integers to be in any
order (ex:a12b3c or 1ab32c etc), would the total numbre be
(26^3*10^3)*6= 105,456,000?
Thanks.
When you replace the 26-letter alphabet with the
36-element set of alphanumerics, the number of
combinations of 6 is easily seen to be 36^6.
--
Rich Ulrich
http://www.pitt.edu/~wpilib/index.html
True but I don't think that's the scenario the OP is describing. The
number plates are constrained to being 3 numbers and 3 letters, albeit
in any order. |
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| Richard Ulrich |
| Posted: Wed Apr 16, 2008 8:31 pm |
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On Wed, 16 Apr 2008 04:53:59 -0700 (PDT), Francogrex <franco@grex.org>
wrote:
Quote: Hi, please check if I'm doing this right: if we have a licence plate
of the type 3 letters-3 integers (ex: abc123) then the total number of
unique plates would be equal to 26^3*10^3= 17,576,000. Right?
But what if we allow the 3 letters and the three integers to be in any
order (ex:a12b3c or 1ab32c etc), would the total numbre be
(26^3*10^3)*6= 105,456,000?
Thanks.
When you replace the 26-letter alphabet with the
36-element set of alphanumerics, the number of
combinations of 6 is easily seen to be 36^6.
--
Rich Ulrich
http://www.pitt.edu/~wpilib/index.html |
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