Question 1083529
You are probably working on hypergeometric distribution.
We will start with a contingency diagram, which breaks down the cats and dogs (C and D), and adults and babies (A, B)
---  C   D  Tot
 A  21  16  37
 B  14  12  26
Tot 35  28  63
 
1. 6 animals chosen from 63. 
P(#kittens>=1)
=1-P(#kittens=0)
={{{(1-49/63*48/62*47/61*46/60*45/59*44/58)}}}
=0.7942
 
But that's not the only way.  Using hypergeometric distribution, the probability 
P(#kittens=0)=C(14,0)*C(49,6)/(63,6)
where C(14,0) is the number of combinations of 0 objects taken from 14.
The formula is basically
C(S,s)*C(F,f)/C(S+F,s+f)
where 
S=number of successes in pool
s=number of successes in sample
F=number of failures in pool
f=number of failures in sample.
Here there are 14 kittens in pool (S=14) and 49 others (F=49)
we are calculating no kittens, so s=0, and f=6.
This evaluates also to 0.7942.
 
We will be using this formula to solve the remaining two problems.
 
2. 8 animals (s+f=8) from 37 adults (S+F=37).
The distribution of cats and dogs is 21:16, so the most likely arrangements could be
6:2, 5:3, 4:4
We'll see.
P(6:2)=C(21,6)*C(16:2)/(37,8)=0.1687
P(5:3)=C(21,5)*C(16:3)/(37,8)=0.2952
P(6:2)=C(21,4)*C(16:4)/(37,8)=0.2822
Which shows that the most probable arrangement is 5 cats and 3 dogs.
 
3. A group of 10 baby animals (out of 14 kittens and 12 puppies) is chosen.  The probability distribution is skewed.  You can show this by showing that P(4:6) does not equal P(6:4), or other similar ratios.