I have three sets of cards -- A , B , C , each consisting of eight cards. I decide to redistribute the cards to form the three sets of eight cards each. How many ways can it be done if no card has all the same teammates as before ? (All $24$ cards are distinct. )
We are allowed to use only inclusion-exclusion.
Can anyone give me any hint?
My attempt : Universal set will be $24!$ . But I can not find the excluded sets. If only $A$ remains intact , but $B$ and $C$ do not. We will get $16! - (8!)^3$ sets. Same will go when $B , C $ remain intact. $A$ , $B$ , $C$ remain intact in $(8!)^3$ ways.
So the required number will be $24! - 3 \times (16! - (8!)^3) - (8!)^3$ .
We can arrange the cards into $3$ groups of $8$ dealing the first $8$ cards into one pile, the next 4 into a second pile, and the remaining $8$ into a third pile. There are $24!$ ways the deck can be arranged originally, and it doesn't matter what order the cards are dealt into a particular pile, so that gives $\frac{24!}{(8!)^3}$. Furthermore, the order of the piles themselves doesn't matter, so there are $$\frac{24!}{(8!)^33!}$$ ways to divide the cards into $3$ groups of $8$.
I divided by $3!$ because there are $3!$ ways to arrange the piles themselves; we don't really care if a group of $8$ cards is in the first pile or the third pile, so long as they're in the same pile.
Now we need to exclude those arrangements that include one of the forbidden piles. There are $3$ such piles. When we choose one, we have to split the remaining $16$ cards into $2$ groups of $8$. Reasoning as above there are $$\frac{16!}{(8!)^22!}$$ ways to do that, which gives $$\frac{24!}{(8!)^33!}-3\frac{16!}{(8!)^22!}$$ ways so far.
Now we still have to worry about double-counting. Suppose two of the forbidden groups are included. Then so is the third. This arrangement has been subtracted $3$ time so we have to add it back in twice, giving a final answer of $$\frac{24!}{(8!)^33!}-3\frac{16!}{(8!)^22!}+2$$