Index of a permutation that has repetition

Question

With 8 of 10 digits (ex: 0,1,2,3,4,5,6,7,8,9) we could have 1360800000 numbers; as you can see calculated here:

Partitions:
one digit appears 3 times, (example: 3749832414)
$\dbinom{8}{1}\dbinom{10}{2}\biggl(\dfrac{10!}{3!}\biggl)= 217728000$
two digits appear 2 times each. (example: 7364512193)
$\dbinom{8}{2}\dbinom{10}{2}\biggl(\dfrac{10!}{2!2!}\biggl)= 1143072000$
and sum of them is: 1360800000

The question is: how to get index of these numbers? For example, what is the index of 3749832414 or 7364512193?
There is a solution in this link, but it doesn't provide an index according to 10 digits, but provides an index according to digits that provided for the function. So it will return 0 for both 0123456777 and 0123456888 which does not answer the need for finding the index for each number if we consider 10 digits.
Edit
Leading zeros are allowed.
It doesn't matter how to index these numbers, just indexing them to be able to know what index represents what number and vise versa.
Edit
1360800000 is the count of numbers that have 8 of 10 digits and length of 10. In others words, all the 10 digit numbers that have two of their digits appeared two times each or one of their digits appeared 3 times like 3749832414 or 7364512193, in other words 1360800000 is count of all the permutations that have 2 global extra appearances of digits (0,1,2,3,4,5,6,7,8,9) in the length of 10 digits/symbols which causes the appearance of 8 digits out of 10 available digits.

Python code to calculate the count of these numbers:

c2 = binomial(8, 1) * binomial(10, 2) * (factorial(10) / factorial(3))
c11 = binomial(8, 2) * binomial(10, 2) * (factorial(10) / (factorial(2) * factorial(2)))
print(c2)
print(c11)
csum = c2 + c11
print("sum is: ", csum)

which prints :

217728000.000000
1143072000.00000
sum is: 1360800000.00000

Answer 1

$\color{red}{\text{Warning!}}$ I am writing this answer to show the mathematical approach , but this is very cumbersome , so i recommend you to use computer program instead of combinatorical approach.

Saying that the first digit can be zero , make our work easier.I think saying "strings" instead of "digits" can be more appropriate in that sense.Anyway , lets turn our attention.I think that the best way to count these numbers is to use exponential generating functions.Otherwise ,it would be so cumbersome to deal with .

As you indicated ,a number can repeat at most $3$ times because there is $8$ distinct number in this string of lenght $10$.Then , the E.G.F of each number can be $$\bigg(x + \frac{x^2}{2!}+\frac{x^3}{3!}\bigg)$$

Now , for $3749832414$, we must firstly count how many string there are such that starting with $0$ , $1$ ,$2$.

• For the strings start with $0$ :

If it starts with zero , then the E.G.F form of zero is $$\bigg(x + \frac{x^2}{2!} \bigg)$$ Now choose the rest $7$ digits by $\binom{9}{7}$.

So ,the number of strings of lenght $10$ starting with zero and containg $8$ distinct numbers is : $$[x^{10}]\bigg(x + \frac{x^2}{2!}+\frac{x^3}{3!}\bigg)^7\bigg(x + \frac{x^2}{2!}\bigg)$$

This calculation is also valid for those strings that start with $1$ or $2$ .So when we come up to the strings starst with $3$ there are $3 \times [x^{10}] \times \binom{9}{7}$ strings.

• For starting with $30$ :

We need to select $6$ numbers among the rest $8$ by $\binom{8}{6}$ , and find $$[x^{10}]\bigg(x + \frac{x^2}{2!}+\frac{x^3}{3!}\bigg)^6\bigg(x + \frac{x^2}{2!}\bigg)^2$$

This is valid for starting with $31,32,34,35,36$ ,so $$6 \times \binom{8}{6}[x^{10}]$$

However ,for those starting with $33$ , the E.G.F is $$[x^{10}]\bigg(x + \frac{x^2}{2!}+\frac{x^3}{3!}\bigg)^7\bigg(x \bigg)$$

Moreover, we must select $7$ numbers instead of $6$ ,so $$\binom{9}{7}[x^{10}]\bigg(x + \frac{x^2}{2!}+\frac{x^3}{3!}\bigg)^7\bigg(x \bigg)$$

Now , you must the number of strings starting with $370,371,372,373$ using the same logic.This process will be applied for each next digits.I hope you understand the logic.

However, this is very cumber some as i said , so i would recommend you to use computer program.