What is the intuition behind the resolution method for this problem?

Question

I'm trying to understand the intuition behind the way we proceed to calculate how many different 4-digit numbers can we build using digits from $1$ to $9$, such that there is exactly one digit repeated, so for example $1982$ or $6333$ are not valid, but $2942$ and $7721$ are. To find the result, we calculate

$$9\cdot \text{C}^4_2 \cdot \text{P}^8_2 = 9\binom{4}{2} \cdot \frac{8!}{6!} = 3024$$

Now when I look at $\text{C}^4_2$ I can't think in other terms than "amount of possible subsets of size 2 that we can build from a set of 4 elements", so I really can't get the intuition of multiplying this by "the amount of different size-2 permutations we can create from a set of eight elements, without repetition". I would appreciate someone explains to me the logic of this operation, because I don't like to do mathematics mechanically and without understanding what I'm doing. Thank you.

Answer 1

• Choose 1 digit out of $1, \ldots, 9$: $\color{blue}{9}$ possibilities
• Choose $2$ places out of $4$ to place this digit twice: $\color{blue}{\binom{4}{2}}$ possibilities
• Fill the remaining $2$ places with $2$ different digits from the remaining $8$ digits: $\color{blue}{8\cdot 7 = P_2^8}$ possibilities

All together $$\color{blue}{9\cdot \binom{4}{2} \cdot 8 \cdot 7}$$