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How many different license plates can fit this description? A_ _ and three digits which contains 1 and 2

  • #1
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Homework Statement:

A witness to a hit-and-run accident tells the police that the license plate of the car in the accident, which contains three letters followed by three digits, starts with the letter A and contains both the digits 1 and 2. How many different license plates can fit this description?

Relevant Equations:

Counting

permutation(?) = P(n,r) = n!/(n-r)!
I have a question and searched about at google and found an answer which I don't make sure. If there is 26 letters and 10 digits;
my answer is:
first letter: 1 way(which is A)
second letter: 26 way
third letter: 26 way
first digit: 1 way(which is 1)
second digit 1 way(which is 2)
third digit: 10 way
So, I think 26x26x10 license plates can fit this description.
But the other guy had said:
in addition above statements: order three digits ; P(3,3), order three letters ; P(3,3) , thus 26x26x10xP(3,3)xP(3,3)

Which one is correct? Would you please tell how we should think?

DarkFuzzy02.jpeg

edit: I add a photo above so new arrivals can know the drill easily .
 
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Answers and Replies

  • #2
kuruman
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"Contains both the digits 1 and 2" does not necessarily mean that the first number is 1 and the second 2. One could also have 1*2, *12, 21*, etc.
 
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  • #3
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Now it is falling into place. I must have missed out some things. Thanks both of you for attention
 
  • #4
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"Contains both the digits 1 and 2" does not necessarily mean that the first number is 1 and the second 2. One could also have 1*2, *12, 21*, etc.
I agree with this. The thread title implies that the digits part of the plate are 12n, but the problem statement says merely that the number part contains the digits 1 and 2.

Also, the formula quoted for permutations is wrong:
P(n,r) = n!/(n-r)! x r!
The correct formula is ##P(n, r) = \frac{n!}{(n - r)!}##
See https://en.wikipedia.org/wiki/Permutation, about a third the way down the page.
 
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  • #5
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I agree with this. The thread title implies that the digits part of the plate are 12n, but the problem statement says merely that the number part contains the digits 1 and 2.

Also, the formula quoted for permutations is wrong:

The correct formula is ##P(n, r) = \frac{n!}{(n - r)!}##
See https://en.wikipedia.org/wiki/Permutation, about a third the way down the page.
Thanks to tell the problem with question tread. I've just edited these.
 
  • #6
berkeman
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Thanks to tell the problem with question tread. I've just edited these.
So can you post your answer? I want to check mine... :wink:
 
  • #7
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So can you post your answer? I want to check mine... :wink:
Of course I can, here below;
WhatsApp Image 2020-05-24 at 01.26.19.jpeg

I hope our answers are the same 😅
 
  • #8
berkeman
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I hope our answers are the same
I processed your dark picture a bit so I could try to read it...

DarkFuzzy02.jpeg


I didn't use any Permutations or Combinations; I only used counting. But I'm not a mathematician, so others will have to chime in about whether your answer is correct. My answer was a bit north of 6 million...
 
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  • #9
etotheipi
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I got the same as @requied's answer for this one
 
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  • #10
DaveC426913
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Nitpick: I can't speak for everywhere, but where I live, there are several letters that never appear on any standard plate:
G, I, O, Q and U.

This reduces the plate alphabet to 21 letters.

It differs from place to place, but I, O and Q are pretty universal in 'Murica, making the alphabet 23 letters.

(Of course, this does not apply to vanity plates).

While that's probably outside the scope of the question - mentioning it, and providing a second solution - might net you bonus marks!
 
  • #11
berkeman
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Nitpick: I can't speak for everywhere, but where I live, there are several letters that never appear on any standard plate:
G, I, O, Q and U.
Interesting, and useful.

I can see O (zero confusion) and maybe I (1 confusion?), but G and U? Inquiring minds want to know... :smile:

Personally, here in Cali, I can't tell the difference between E and F if the license plate frame comes up too much.
 
  • #12
DaveC426913
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Interesting, and useful.

I can see 0 (zero confusion) and maybe I (1 confusion?), but G and U? Inquiring minds want to know... :smile:
6 and V.
 
  • #13
DaveC426913
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1590274130519.png

The next day: 'What? Six bank robberies!? But I just vandalized the library!' 'Nice try. They saw your plate with all the 1's and I's.' 'That's impossible! I've been with my car the whole ti-- ... wait. Ok, wow, that was clever of her.'

https://xkcd.com/1105/
 
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  • #14
berkeman
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I got the same as @requied's answer for this one
Okay, after a quick PM conversation with @etotheipi I see what I did wrong to overcount up to 6 million. I agree now. Thanks! :smile:
 
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  • #15
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I got the same as @requied's answer for this one
So did I.
If we look at just the part with the numbers, there are P(3, 3) = 6 permutations of the symbols 1, 2, *. The * symbol can be any digit 0, 1, 2, …, 9, so for each of the 6 permutations there are 10 possible number patterns. In total, there are 60 different number patterns.

Combining this with the ##26^2## possible letter patterns A**, we have ##26^2 \cdot 60## = 40, 560 different license plates.
 
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  • #16
DaveC426913
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Hmph. 60 x 232 = 31,740. :wink:
 
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  • #18
berkeman
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So, math may change from place to place o0)
Yup, but only places with less constrained license plates, like places that allow custom license plates. If the plate is constrained absolutely not to mix alpha and digits, it's still 26*n...
 
  • #19
kuruman
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So, math may change from place to place o0)
And to a place where genuine Arabic numerals are used ...

Arabic .jpg
 
  • #20
DaveC426913
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So, math may change from place to place o0)
Well...
The fact that it does not specify a state means any letters that are allowed in even one location, should be counted. But if there are letters that are never allowed, then that would reduce the number - which is theoretically fixed and unilateral.

But, technically, it's negated by the existence of vanity plates (assuming they are exempt from the IOQ rules).
 
  • #21
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I just did a little research and saw that in Chinese there is not letters, but characters around 4000 :) I wonder how they set therir license plates :D
 
  • #22
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Hmph. 60 x 232 = 31,740. :wink:
Where are you getting 23?
Each of the two missing letters can be any of the 26 letters of the alphabet.

I still stand by the OP's and my value, 40,560.
 
  • #23
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Where are you getting 23?

It differs from place to place, but I, O and Q are pretty universal in 'Murica, making the alphabet 23 letters.
I think he is thinking about a different place and just kidding.
 
  • #24
berkeman
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Where are you getting 23?
Each of the two missing letters can be any of the 26 letters of the alphabet.

I still stand by the OP's and my value, 40,560.
Yes, OP's and yours and @etotheipi answers are correct. Dave is bringing up the fact that for real license plates, there are other considerations. A little off-topic, but a useful real-world side comment IMO once the problem is solved.

Related -- we have to limit the characters that are allowed in our default passwords on some of our Ethernet-connected products. Our users too often misread the default passwords on our product labels and cannot use our products when they receive them (and we recommend changing the password right away). "I" and "1" are one example...
 
  • #25
DaveC426913
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I think he is thinking about a different place and just kidding.
Part kidding, more pedantic. Maybe useful.

The problem does not say anything about what letter set is expected. It is an assumption made by we as problem solvers that all letters of the alphabet are valid. It's a fair assumption, but an assumption nonetheless.
But the problem does specify a context (license plates) - one in which a non-conventional letter set is specified.


It's the kind of thing that could earn the student bonus marks for pointing out ( definitely in addition to providing the expected answer!)
 

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