Elven mystical numbers

Generalization

Let's find a general formula for the number of $n$ -digit EMN s, where $n$ is an odd positive integer.

The properties of EMN imply that the set of all $n$ -digit EMN s forms a bijection with the set of all $\dfrac{n+1}{2}$ -digit integers
whose digits are in non-decreasing order and the last digit is the largest digit. Thus, it is sufficient to find the number of latter ones.

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Again, a new bijection can be formed because such $\dfrac{n+1}{2}$ -digit integers can be represented as sequences of $\dfrac{n+1}{2}$ # s and 9 + s, where # s are the numbers and number of + before a particular # is the digit which is in that place. For instance $125$ is the same as +#+#+++#++++ .

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The problem is now the same as finding the number of ways to arrange 9 + s and $\dfrac{n+1}{2}$ # s, where $2$ of the + s are fixed (only $1$ is fixed if $n = 1$ ) to ensure that the first digit is at least $1$ and that the last digit is larger than any other digit.

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Now we have to find the number of arrangements of 7 + s and $\dfrac{n+1}{2}$ # s
which can easily be calculated to be $\displaystyle{{\dfrac{n+15}{2}} \choose {7}}$ , when $n > 1$ using Stars and Bars .

If $n = 1$ , the number of EMN s is $1$ more than that.

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Answer

Plugging $n = 11$ to the general formula gives the answer of $\displaystyle{{13} \choose 7} = \boxed{1716}$ .

Relevant wiki: Combinations - Problem Solving

Solution:

One way is to calculate for each middle digit (peak), so if $9$ is the middle digit, how many combinations for other digits are there and so on....

Second way: On how many ways can 6 digits be chosen? The order in which numbers are selected is irrelevant, because after selection, they will be sorted (so the rule (3) is fulfilled). So 6 digits can be chosen on $14C6$ ways, $14C6 = \frac{14!}{6! \times 8!}$ [1]. To satisfy rule (2), it is needed to subtract all the combinations in which the greatest number is chosen more than once.

If 5 digits are chosen, after which the greatest one of them is added once, the result is 6 digits where the greatest one definitely appears more than once. The number of these unsatisfying combination is $13C5 = \frac{13!}{5! \times 8!}$ [2]

When [1] is subtracted by [2], the number of satisfying combinations is $\boxed{1716}$

The integral equation approach.