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The problem asks how many different six digit license plates can be made if there are three digits that occur once and one digit that occurs thrice?

My idea is that there can be 10 numbers for the reoccurring digits, and $6 \choose 3$ ways to lay them out. Then there are $9\cdot8\cdot7$ ways to lay out the remaining unique digits. This gives $10\cdot{6\choose3}\cdot9\cdot8\cdot7$ possible plates.

Is this thought process correct? I'm having some trouble on these problems.

Thanks

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  • $\begingroup$ Why do you have $9! \cdot 8 \cdot 7!$? What does $x$ factorial mean? $\endgroup$
    – XRBtoTheMOON
    Oct 16, 2017 at 15:04
  • $\begingroup$ Just to be clear ... you have one digit that appears three times, and three other digits that are all different? You may want to have the problem statement fully defined in the body of your post. Having it stated between the title the the body is a bit confusing. $\endgroup$
    – Bram28
    Oct 16, 2017 at 15:05
  • $\begingroup$ Sorry, I just fixed the statement. $\endgroup$
    – user402326
    Oct 16, 2017 at 15:09

2 Answers 2

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Then there are $9\cdot8\cdot7$ ways to lay out the remaining unique digits.

This is not about the number of ways of 'laying them out', but of picking digits for the three remaining digits.

So yes, you have indeed $6 \choose 3$ ways to lay out the recurring digit, and you have $10$ choices for what that digit is.

But then for the other three digits, you can pick $9$ choices for the 'first' (i.e. the first one you encounter from left to right in the digit string) digit, $8$ for the next, and $7$ for the last, giving you a total of:

$$10 \cdot {6 \choose 3} \cdot 9 \cdot 8 \cdot 7$$

Alternatively:

There are $9 \choose 3$ ways to pick the $3$ remaining digits, and $3!$ ways to lay them out:

$$10 \cdot {6 \choose 3} \cdot {9 \choose 3} \cdot 3!$$

which works out to be the same (of course!):

$$10 \cdot {6 \choose 3} \cdot {9 \choose 3} \cdot 3! =$$

$$ 10 \cdot {6 \choose 3} \cdot \frac{9!}{3!\cdot 6!} \cdot 3! =$$

$$10 \cdot {6 \choose 3} \cdot \frac{9!}{6!} =$$

$$10 \cdot {6 \choose 3} \cdot 9 \cdot 8 \cdot 7$$

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  • $\begingroup$ Thanks, I must have fixed the statement while you were typing. It was just a typo $\endgroup$
    – user402326
    Oct 16, 2017 at 15:12
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    $\begingroup$ @sadlyfe OK, then you had it all ok!! ... though again the $9\cdot 8 \cdot 7$ term isn't about 'laying them out' but about picking the three remaining digits. So phrase that a bit differently, and you're all good! :) $\endgroup$
    – Bram28
    Oct 16, 2017 at 15:16
  • $\begingroup$ Ok, that makes sense! Thanks $\endgroup$
    – user402326
    Oct 16, 2017 at 15:18
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How you handled the recurring digit is correct. The issue was how you initially counted the arrangements of the remaining digits.

There are $9$ choices for the first unique digit slot, $8$ for the second, and $7$ for the third. So the total number of choices would be

$$N = 10 \cdot {6 \choose 3} \cdot 9 \cdot 8 \cdot 7.$$

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  • $\begingroup$ Sorry, I had a few typos and I fixed them. Is what you just said the same as my fixed answer? Or are you saying I should remove the 10. If I should remove the 10, then why? $\endgroup$
    – user402326
    Oct 16, 2017 at 15:09
  • $\begingroup$ Edited my answer. $\endgroup$
    – John
    Oct 16, 2017 at 15:17

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