The problem is "how many ways can we distribute $12$ unique cards to $4$ persons with every person receiving at least $1$ card"
Without the requirement that everyone getting at least $1$, it would be $4^{12}$, and my initial thought was that with the req. It would be $_nP_r(12,4)\times 4^8$ but that's way too high, looks like I need to divide with some permutations that's already accounted for..
Please help!
On
I think you can solve this directly. First you can consider the number of way to give 12 cards to 4 players with every player having at least one card and then multiply by the number of permutation of the cards (that is $12!$).
Let $n_i$ be the number of cards of the $i^th$ player. You can have $$1\leq n_1\leq 9$$ because you have to keep at least $3$ cards for the other players. Then, for the second player, it comes $$1\leq n_2\leq 10-n_1.$$ Finally for the third we have $$1\leq n_3 \leq 11-n_1-n_2.$$ Note that for the fourth player, his number of cards is fixed by the previous three. In total, we have $$\sum_{n_1=1}^9(10-n_1)\left(\sum_{n_2=1}^{10-n_1}(11-n_1-n_2)\right)=\sum_{n_1=1}^9(10-n_1)\left((10-n_1)(11-n_1)-\sum_{n_2=1}^{10-n_1}n_2\right).$$ I let you finish the computation (and don't forget to multiply your result by $12!$).
On
As you observed, without restrictions, the $12$ cards can be distributed to $4$ people in $4^{12}$ ways. From these, we must exclude those distributions in which fewer than four people receive a card.
There are $\binom{4}{1}$ ways to exclude one person from receiving a card and $3^{12}$ ways to distribute the cards among the remaining three people. Hence, there are $\binom{4}{1}3^{12}$ ways to exclude one person from receiving the cards.
However, if we subtract $\binom{4}{1}3^{12}$ from $4^{12}$, we will have subtracted too much since we will have subtracted those distributions in which two people do not receive a card twice, once for each way of designating one of those people as the person who does not receive a card. Since we only want to exclude them once, we must add such distributions back.
There are $\binom{4}{2}$ ways to exclude two people from receiving a card and $2^{12}$ ways to distribute the cards among the remaining two people. Thus, there are $\binom{4}{2}2^{12}$ ways to distribute the cards so that two people are excluded from receiving a card.
However, when we subtract distributions in which one person does not receive a card and then add distributions in which two people do not receive a card, we first subtract distributions in which three people do not receive a card three times, once for each of the three ways of designating one of those three people as the person who does not receive a card, and then add them three times, once for each of the $\binom{3}{2}$ ways of designating two of those three people as the people who do not receive a card. Consequently, we have not excluded such distributions at all. Therefore, we need to subtract them.
There are $\binom{4}{3}$ ways to exclude three people from receiving a card and one way to give the remaining person all the cards.
Hence, by the Inclusion-Exclusion Principle, there are $$4^{12} - \binom{4}{1}3^{12} + \binom{4}{2}2^{12} - \binom{4}{3}1^{12}$$ ways to distribute the $12$ cards to four people so that each person receives at least one card.
Each admissible allocation is a surjective map $f:\>[12]\to[4]$. Such maps are counted by the Stirling numbers of the second kind. To be precise: The number of surjective maps $f:\>[n]\to[k]$ is given by $k!\, S(n,k)$. There are recursions and tables for these numbers, but no closed formula in terms of factorials etc.