A set of playing cards consists of 36 cards. These 36 cars are subdivided into 9 types (A,B,C, (...)I) with four cards each. The cards within these subtypes do not differ (A1=A2=A3=A4). How many possibilites are there, to take up 9 cards? The order of taking up cards and the order of the taken up cards do not matter (AAABBCCCD=ABABACDCC).

I somehow do not get a correct result. First i thought of how man possibilites there are to take 9 cards from 36, but i dont know how to get rid of all possibilites. I tried 36!/27!4!^9, but this is totally wrong...

thanks for your help

Here is a feasible but somewhat laborious way of doing it. Consider, for example, the case where you are given 4 cards from one of the 9 subtypes, 3 from another, and 1 from yet two other subtypes. How many cases of that general sort, "4+3+1+1" are there?

Well, you can choose the subtype that contributes 4 cards in 9 ways. Having choosen that subtype, you can choose the subtype that contributes 3 cards in 8 ways, and the two subtypes that contribute 1 card each can be chosen in \(\displaystyle \binom{7}{2}\) ways. In all, there are \(\displaystyle 9\cdot 8\cdot\binom{7}{2}=1512\) different cases of being dealt 9 cards that follow this general "4+3+1+1" pattern.

So, I think, it comes down to listing all these possible ways of partitioning the number 9 into a sum of at most 9 non-negative integers (\(\displaystyle \leq 4\)) and, for each of these partitionings, determining how many ways there are to get them by combining the 9 subtypes of cards accordingly.

Such a list of partitionings might start like this (I am leaving out trailing 0s)

\(\displaystyle \begin{array}{lcl}9 &=&4+4+1\\

&=& 4+3+2\\

&=& 4+3+1+1\\

&=& 4+2+2+1\\

&\vdots& \vdots

\end{array}\)

To keept this list reasonably small, you should only list the sums that have non-increasing terms (from left to right).