So we have $2$ cats, $2$ dogs, $2$ lions, $2$ tigers, and we want to split them into $4$ groups such that each group has $2$ none similar animals.
In how many ways we can do that?
I'm trying to use the inclusion exclusion principle, since I tried to approach it normally and got stuck, and here's what I reached:
numbers of ways with $1$ group with $2$ similar animals : $8 \choose 2$$\frac {6!}{2!2!2!}$ .
number of ways with $2$ groups with $2$ similar animals: $8 \choose 4$$\frac{4!}{2!2!}$.
number of ways with $3$ groups with $2$ similar animals: $8 \choose 6$.
number of ways with $4$ groups with $2$ similar animals: $8 \choose 8$.
So my answer will be $\frac {8!}{2!^4}$ – $4 \choose 1$$8 \choose 2$$\frac {6!}{2!2!2!}$ + $4 \choose 2$$8 \choose 4$$\frac{4!}{2!2!}$ – $4 \choose 3$$8 \choose 6$ +$1$.
$=5040 – 10080 + 2520 – …= (-)$.
So basically I got it all wrong since I got a minus in the end, would be happy to see where are my mistakes and how to solve this. Thanks in advance.
Best Answer
There are essentially two different patterns: either you have two species in two groups (forcing the choice for the other two groups) or you do not (in which case each group is distinct and with four groups you must have a cycle):
(a) AB AB CD CD
(b) AB BC CD DA
But the next issue is what counts as a different split.
If the species are distinguishable, but the groups and individual animals are not then (a) has $3$ possibilities (the cats can be with one other species) and (b) has $3$ possibilities (the cats can not be with one other species), making $6$ in total
If the species and individual animals are distinguishable, but the groups are not then (a) has $3\times 2^2$ possibilities (since you can only split a pair of species two ways, but have two sets of pairs) and (b) has $3\times 2^4$ possibilities (since going round the cycle you get two choices for each species), making $60$ in total