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This is a question about license plates that for the life of me I can't figure out why it's wrong. The question is: A license plate have 3 letters and 3 numbers. What are the total number of possible license plates if there is no restriction on which order letters and numbers appear? My answer was:

$10^3 \cdot 26^3 \cdot 6!$. The answer however says that in place of 6! it's

$10^3\cdot 26^3 \cdot \binom 6 3$ instead. I can't understand why it's not $6!$, my reasoning is that I treat each number and letter as its own entity and they can be arranged in $6!$ ways?

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Hint: simplify the problem - suppose that license plates consist of just two numbers. By the logic you've put forth, there should be $10^2 \cdot 2! = 200$ possible license plates. But we know that's wrong - there's only 100 two-digit numbers. What went wrong? – senshin May 4 '14 at 4:01
Hmm that's a great example. Is it that the order doesn't matter since 00 is the same regardless of when the 0's arrive? In my example then why is 6 choose 3 and why don't we just divide by 2? – Roger May 4 '14 at 4:06

You choose which of the three positions will have letters. There are $\binom 6 3 = 20$ ways to do that. Then you choose the three digits and the three letters. There are $26^3 10^3$ ways to do that.

If you first choose three digits and three letters and then decide to shuffle them, it might seem plausible that there are $10^3\cdot26^3\cdot6!$ ways to do that. But suppose you choose $ABB262$. There are $6!=720$ orders in which to place six distinguishable objects. But these are not distinguishable: suppose you interchange the two $2$s and also the two $B$s: you get a different item in the list of $720$, but it should be counted as the same item in the list that you want. There are many similar cases, so you get a substantial overcount.

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In your answer, $10^3$ is the number of ways to choose $3$ digits in a specific order, and $26^3$ is the number of ways to choose $3$ letters in a specific order. Then $6!$ is the number of ways to determine the order of these $6$ symbols.

Do you see the error? - you have chosen the order of your digits twice, once in the first step and once in the third (and likewise for the letters). But if the order of the digits has already been settled in the first step, then many of the $6!$ orderings in the third step are no longer possible, and the number of orderings in the third step should actually be considerably fewer.

IMO, usually the easiest way to solve this kind of problem, where you have to choose from different kinds of objects such as letters and digits, is to choose the placement of the two kinds first, and only then the specific objects. So I would do it like this.

  • Choose which of the six places are to be occupied by digits. Note that (i) we cannot choose a place more than once - for example, it does not make sense to say that our three digits will be in places $1,1,5$; (ii) the order is irrelevant - as we have not yet chosen the specific digits, it does not matter whether we say they will be in places $1,2,5$ or $1,5,2$. Therefore we have to choose $3$ places from $6$, with repetition forbidden and order not important: this is a combination, and the number of possibilities is $\binom{6}{3}$.
  • Now choose the three digits - repetition is OK and order is important, so the number of ways is $10^3$ as in your answer.
  • Then choose the three letters - the number of ways is $26^3$.

So the total number of licence plates is $\binom63\times10^3\times26^3$.

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Let's pretend that the letters and numbers are the faces that land up when you roll 6 dice (3 of them have 10 sides, call then $N_1$, $N_2$, and $N_3$, and 3 of them have 26 sides, calkl them $L_1$, $L_2$, and $L_3$). When you roll the die but don't record the faces yet, there will be $10^3\cdot 26^3$ possibilities. Now you try to write down the result and order matters. Where do you write the faces of the, say, three letter-dice $L_1$, $L_2$, $L_3$? You have 6 choices and from these you choose three.

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We concentrate on one of the ways in which $10^3\cdot 26^3\cdot 6!$ is not correct. We can think of $10^3\cdot 26^3$ as the number of ways to get a string of $3$ digits followed by a string of $3$ letters. So 127BKD is among the strings already counted. But 172KBD has also already been counted, as has 217DBK, and several others. Thus the permutations of 127BKD that leave us with digits in front and letters after would be multiply counted if we multiplied $10^3\cdot 26^3$ by $6!$.

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