Let's consider taking each animal in turn so that, in order, we have PPDDCC. We can estimate the probability of getting this permutation. P=pig, D=dog, C=cat.
4 pigs in 15 animals makes the probability of selecting a pig 4/15, leaving 14 animals, 3 of which are pigs.
The second animal is also a pig, probability of 3/14, leaving 13 animals.
3rd animal is a dog, probability of 5/13, 4th is a dog, probability of 4/12, leaving 11 animals.
5th animal is a cat, probability 6/11, last is a cat, probability of 5/10.
The probability of this particular permutation is the product of the individual probabilities:
(4/15)(3/14)(5/13)(4/12)(6/11)(5/10). The product of the denominators is 15P6=3603600.
The product of the numerators is 12×20×30=7200. So we have 7200/3603600=2/1001.
Now consider another permutation: DDCCPP
The combined probability is:
(5/15)(4/14)(6/13)(5/12)(4/11)(3/10).
This is exactly the same product as for PPDDCC. We have the same set of numerators and denominators. So, no matter what order the pets are selected we still have the same probability of 2/1001=0.002 approx or 0.2%.
This is the same as 4P25P26P2/15P6.