All Unique Three Digit Combinations using four 8s, three 9s, seven 1's and three 5s

Question

So basically I have find the unique possible combinations of 3 digit number using four 8s, three 9s, seven 1's and three 5s, e.g.
"888"
"819"
"891"
"855", ..., etc
Because of having same number present multiple times I am not able to determine the answer. Some of the approaches tried were $\frac{17!}{4!3!7!3!}$. All answers are appreciated.
Thanks

Answer 1

In general, choose "numbers of kinds" and permute

$3\,$ of a kind: $\;\binom41\times\frac{3!}{3!}$

$2-1\,$ of a kind: $\;\binom41\binom31\times\frac{3!}{2!1!}$

$1-1-1\, $ of a kind: $\;\binom43\times \frac{3!}{1!1!1!}$

Compute and add up

But here, $\geq 3$ digits are available for each digit, so simply $4^3 = 64$ possibilities

Answer 2

These types of questions can be easily solved by exponential generating functions. It will give you power.

It is said that there are four $8's$ , three $9's$ , seven $1's$ , and three $5's$ .In other words , we have $8,8,8,8,9,9,9,1,1,1,1,1,1,1,5,5,5.$

We want to find the number of all possible $3$ digits numbers using these digits without any restriction. So , we should firstly write exponential generating function forms of each digits.

Exponential generating function of $8$ : $$\bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\bigg)$$

Exponential generating function of $9$ : $$\bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} \bigg)$$

Exponential generating function of $1$ : $$\bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} +\frac{x^6}{6!} +\frac{x^7}{7!}\bigg)$$

Exponential generating function of $5$ : $$\bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} \bigg)$$

Now , to find the number of all possible $3$ digits , we should find the expansion of these exponential generating functions such that $$\bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\bigg) \times \bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} \bigg)^2 \times \bigg(1+ \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} +\frac{x^6}{6!} +\frac{x^7}{7!}\bigg) $$

After that , we should find the coefficient of $\frac{x^3}{3!}$ or find $x^3$ and multiply it by $3!$. It will give you the result.

Moreover , you can find the number of all possible arrangements of any lenght using this method . For example , if you wnder about lenght $5$ , then we should find the coefficient of $\frac{x^5}{5!}$ or find $x^5$ and multiply it by $5!$.