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So I have been stuck on this question for a few days now and I am honestly just floored at how to get the correct answer. The question is as follows:

A license plate number consists of eight characters. The first two characters are a $2$-character code representing one of $39$ regions. The next three characters are digits $0,1, \ldots,9$, and the last three characters are capital letters A, B, $\ldots$, Z. How many license plate numbers contain the digit 4 exactly once in the string of $3$ digits (after the region code)?

My initial thought is that the first and last three "spots" of the string do not change, so the value will be some multiple of:

$39 \cdot 26^3$.

I am extremely confused as to how to approach the digits that are found in the middle of this question, as my initial logic of generating the value

$9 \cdot 9 \cdot 1$ (4 has to be used only once, when it is used it is removed from the set of digits)

but my professor claims that this does not quite account for the ORDER (??) of the digits. Any explanation of where I am going wrong? Any help would be greatly appreciated.

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  • $\begingroup$ Sorry for butchering some of the grammar in the question. If it needs to be more legible, I can re-post it. I am just very stressed out over this question at the moment, it can be hard to write it all out coherently. $\endgroup$ Jun 21, 2021 at 7:37
  • $\begingroup$ "A license plate number consists of two characters." Is this correct? $\endgroup$
    – user338955
    Jun 21, 2021 at 7:44
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    $\begingroup$ It should be $3\cdot 9\cdot 9$ instead of $1\cdot 9\cdot 9$ because $4$ can be placed first, second or third. $\endgroup$
    – Asher2211
    Jun 21, 2021 at 7:45
  • $\begingroup$ Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. $\endgroup$ Jun 21, 2021 at 9:07

2 Answers 2

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The number $4$ can appear in one of the $3$ locations.

Hence you need to first choose the location first, for the remaining two slots, you have $9 \cdot 9 = 81$ options.

Hence in total: $39\cdot 26^3 \cdot 3 \cdot 81 = 3\cdot 13\cdot 2^3 \cdot 13^3\cdot 3^5=2^3\cdot 3^6\cdot 13^4$

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  • $\begingroup$ Your answer is very much appreciated. There were a few other references that I looked over seeing that the rearrangement of the number 4 in three unique locations was a function of counting sets, which I completely missed. $\endgroup$ Jun 22, 2021 at 18:25
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Answer: The simple combination 3 * 81 * 39 * (26)^3 will generate the correct answer for this question. As a numeral, it is 166567752. Credit to Siong Thye Goh for the help!

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