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Can somebody please explain why don't we count the arrangements of $7$ (digits and letters) as $7!$? In my understanding, it will be $7!\cdot 26^3 \cdot 10^4$ I know it's wrong and the solution tells us $35$ possible letters arrangements and not $7!$ but I can't understand why.

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  • $\begingroup$ 1. You choose the positions where letters will be ${7\choose 3}$ 2. You choose the letters (as you didn't count positions order, you choose left to right) $26^3$ 3. And then digits $10^4$. Is that it? $\endgroup$ May 14 '20 at 18:41
  • $\begingroup$ Because after you placed letters, the positions of both numbers and letters are defined (i.e. you do not have a choice where to place numbers) $\endgroup$
    – Vasya
    May 14 '20 at 18:47
  • $\begingroup$ Compare this to the problem of finding how many different $7$-place license plates are possible when all seven of the entries are very specifically the digit $1$. You should be able to immediately tell that there is one and only one possibility: 1111111. Your erroneous approach would have you think it was $7!\cdot 1^7$ possibilities. $\endgroup$
    – JMoravitz
    May 14 '20 at 18:53
  • $\begingroup$ The $26^3$ and $10^4$ ALREADY order the $3$ letters and order the $4$ digits. All that's left is to specify which of the $7$ positions hold the $3$ letters (and then the other $4$ automatically get the digits). That's $C(7,3)=35$. $\endgroup$
    – Ned
    May 14 '20 at 21:10
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Let's assume the license plate has the 3 letters on the left hand side in a row, and the four letters on the right hand side in a row, like this:

$$\mathrm{AAA}1234$$

We have have 26 possible letters to choose from, and we make this choice three times. So this gives us $26^3$ possible letter combinations. Next, we have 10 single-digit numbers to choose from, and we make this choice 4 times. This gives us $10^4$ possible number combinations. Multiplying these gives us $26^{3}\times 10^4$ possible plates.

On the other hand, if there is no restriction on the placement of letters and numbers, we could have something like this: $$\mathrm{A1A2A34B}$$ But we can still only have 3 letters and 4 numbers.

So, to start out let's decide where to put our three letters. We need to pick 3 places out of 7, so thats $7 \choose 3$ possibilities. Now that we have decided where to put our three letters, the other four spots must be numbers. At this point we have returned to our original problem where we know the positions of the letters and numbers. So the final answer should be $${7 \choose3} \times 26^3 \times 10^4$$

I'm not sure where the $7!$ comes from in your solution, but if you add in an explanation for how you came up with that, I'll be sure to explain why it doesn't work.

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  • $\begingroup$ Thank you so much for a brilliant explanation, I finally got it. 7! got stuck in my mind from a previous exercise when I had to calculate # of 7 different books arrangements and I forgot about the repetitive orders. $\endgroup$ May 14 '20 at 21:51
  • $\begingroup$ Glad I could help. $\endgroup$
    – Owen
    May 14 '20 at 22:10

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