How many combinations can be made from this scenario? I know there are similar questions posted on here, but since their are basically infinite problems like this, Idk how to apply them in certain situations. I have added the question and what I believe to be the answers, but I'm unsure of a few of them if someone could please provide feedback. Thanks!!
Scenario: Determine the number of ways to select 6 animals from a a variety of 20 (different) animals:
1.) If all 6 selected are of different types
$\binom{20}{6}$
2.) If half of them are of one type and the other half of another type
$\binom{20}{1} + \binom{19}{1}$?
3.) If all are the same
$\binom{20}{1}$?
4.) If no conditions are applied
$\binom{20}{6}$?
*note: order does not matter
Thanks again!
 A: We assume that there are $20$ different types of animal, that there are at least $6$ animals of each type, and that animals of the same type, such as agoutis, are indistinguishable. 
1) We need to choose $6$ distinct types. There are $\binom{20}{6}$ ways to do that.
2)  We need to choose $2$ distinct types from the $20$.
3)  We need to choose $1$ type. Easy!
4)  This one is substantially harder! List the types of animal alphabetically. Now consider the equation $x_1+x_2+\cdots+x_{20}=6$. We claim that there is a one-to-one correspondence between solutions of this equation in non-negative integers and choices of animals. If we choose $a_i$ animals of type $i$, that corresponds to the solution $a_1+a_2+\cdots+a_{20}=6$. 
Finding the number of solutions of this equation is a standard "Stars and Bars" problem (please see wikipedia). There are $\binom{20+6-1}{6-1}$ solutions. 
A: (1) We choose 6 animals out of 20, and we don't care about the order. (If someone asks you about the six animals you've selected, he/she won't care if the zebras or the elephants were selected first.). $\binom{20}{6}$
(2) We have to choose two types of animals. Again, which one we choose first doesn't matter: $\binom{20}{2}$
(3) They are all the same (of one type): $\binom{20}{1} = 20$ different options.
(4) There are several different cases here. We might have just one type of animal, for example six zebras. Another way could be that we have zebras and horses, 3 of each. Or 4 zebras and 2 horses. Or in general up to 6 types of animals, but exactly 6 animals chosen in total.
There were 20 different types of animals. If we call the number of different animals that we choose $x_1, x_2, \dots, x_{20}$, and there have to be 6 in total, we can say:
$$x_1 + x_2 + \dots + x_{20} = 6$$
Since we cannot select a negative amount of one type, we have $x_1 \geq 0, x_2 \geq 0, \dots$ etc. This type of problem is solved with the very common combinatorial tool commonly called stars and bars, and the solution is:
$$\binom{20+6-1}{6-1}$$
