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What is a good way to approach this question?

How many $14$-digit even numbers can be formed using $0,1,1,2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5$?

I know that I can list the possible outcomes with different ending numbers and then add up all the outcomes. However, is there a way to solve the question without categorizing adding different types together?

N. F. Taussig
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Kevin
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  • Since the first digit cannot be zero and the number of times each even digit appears is not the same, case work looks like the way to go. Have you obtained an answer doing case work? – N. F. Taussig Sep 11 '19 at 00:47
  • But wouldn't case be an unsustainable way of solving this kind of question? If numbers get more, it would only become harder for case method to work and that is why I want to find a way around for the question. – Kevin Sep 11 '19 at 01:02
  • The requirement that the number is even means there are at most five cases to handle. – N. F. Taussig Sep 11 '19 at 08:40
  • A problem that could be handled using a symmetry argument is finding the number of three-digit positive integers with distinct digits. By cases: if $0$ is the units digit, we can choose the hundreds digit in $9$ ways and the tens digit in $8$ ways; if $0$ is not the units digit, we can choose the units digit in $9$ ways, the hundreds digit in $8$ ways, and the tens digit in $8$ ways. Therefore, there are $1 \cdot 9 \cdot 8 + 9 \cdot 8 \cdot 8 = 72 + 576 = 648$ three-digit positive integers with distinct digits. – N. F. Taussig Sep 11 '19 at 08:41
  • If we ignore the requirement that the hundreds digit cannot be $0$, we can choose the units digit in $10$ ways, the hundreds digit in $9$ ways, and the tens digit in $8$ ways, giving us $10 \cdot 9 \cdot 8 = 720$ strings of three distinct digits. By symmetry, $9/10$ of these do not have a zero in the hundreds place, so there are $\frac{9}{10} \cdot 720 = 648$ three-digit positive integers with distinct digits. – N. F. Taussig Sep 11 '19 at 08:44
  • However, if you try to do the same thing with three-digit positive even integers with distinct digits, the symmetry argument fails because there are more choices for the hundreds digit if $0$ is the units digit than there are if another even number is the units digit. The situation here is analogous. – N. F. Taussig Sep 11 '19 at 08:47

3 Answers3

5

The answer can be written as

$${6\cdot13!-5\cdot12!\over1!2!2!3!3!3!}$$

which computes to $40{,}471{,}200$, as in EuxhenH's answer. The logic here is to start by making the $14$ digits all distinct by temporarily adding subscripts (e.g., the three $5$'s become $5_1$, $5_2$, and $5_3$). The $6\cdot13!$ counts the number of permutations that end with one of the $6$ even digits ($0$, $2_1$, $2_2$, $4_1$, $4_2$, and $4_3$), but this includes permutations that start with the $0$, so we subtract the $5\cdot12!$ permutations that start with the $0$ and end with one of the $5$ other even digits. Finally, we divide by the rearrangements of the copies of the digits, so as to remove the subscripts that we attached to make them distinct.

Barry Cipra
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Case 1: The number ends in $0$. Considering all possible orderings of the other $13$ numbers we get $$\frac{13!}{2!2!3!3!3!}$$ possible outcomes (since we have two of $1$ and $2$ each and three of $3,4,5$ each, we divide by $2!2!3!3!3!$) to remove duplicates.

Case 2: The number does not end in $0$. We have $5$ possible choices for the last digit, which leaves us with $12$ possible choices for the leading digit, $12$ possible choices for the second digit (include zero), $11$ choices for the third and so on. Basically, $$\frac{5\cdot 12\cdot 12!}{2!2!3!3!3!}.$$ Again, we divide to avoid duplicates.

Hence, we get a total of $$\frac{13! + 5\cdot 12\cdot 12!}{2!2!3!3!3!} = 40471200.$$

EuxhenH
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  • @N.F.Taussig, the requirement is not being overlooked. The reference in Case 2 to $5$ possible choices is to the digits $2,2,4,4,4$, not to the digits $1,2,3,4,5$. – Barry Cipra Sep 11 '19 at 11:39
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There are three even numbers, so they make the three cases.

Case 1. Last digit is $0$: $$\frac{13!}{(2!)^2(3!)^3}$$ Case 2. Last digit is $2$: $$\frac{13!-12!}{2!(3!)^3}$$ Case 3. Last digit is 4: $$\frac{13!-12!}{(2!)^3(3!)^2}$$ Hence, adding the three results in $40{,}471{,}200$.

farruhota
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