how many 10 digit numbers contain exactly six 0's and four 1's? how many 10 digit numbers contain exactly six 0's and four 1's?
10 choose 6 or 10 choose 4
how many 4 digit numbers contain at least two 0's?
If there are two zeros then there would be 9x9 
If there are three zeros then there would be 9
If there are four zeros then there is only 1
So would the answer be 9x9 + 9 + 1 ?
how many 4 digit numbers contain at least two 0's or at least two 1's (or both)?
Would it be the same as the last problem but multiply the answer by 2?
 A: Regarding your original question (before you drastically altered it): "How many 10 digit numbers contain exactly six 0's and four 1's?":
That's exactly the same as asking "How many 4-element subsets can I make from the 10-element set $\{10^{9}, 10^{8} , ... , 10^{1}, 10^{0}\}$?". And the answer to that is comb(10,4) = $\frac{10!}{6!4!}$ = 210.
It's also the same as asking "How many 6-element subsets can I make from the 10-element set $\{10^{9}, 10^{8} , ... , 10^{1}, 10^{0}\}$?" But the answer is the same, because comb(n,(n-k)) = comb(n,k) for all n and k, because Pascal's triangle is symmetrical around its central vertical axis. 
As for the remaining questions you later "tacked onto" your original question:
"How many 4 digit numbers contain at least two 0's?"
and
"How many 4 digit numbers contain at least two 0's or at least two 1's (or both)?"
I'll pass. You're not asking "a question", but many. This is starting to look like exactly the same kind of "do my homework for me" questions that plague Quora horribly. I don't answer those, because that's counter-productive. I suggest at least TRYING to do your OWN homework first, then if you get stuck, get on Stack Exchange or Quora and ask SPECIFIC questions (one question per question, not a bunch of questions masquerading as one). 
A: hints
For the first you know the entire number is made up of six ones and four zeros... it's just a matter of placement. You have ten spots and your number will be uniquely specified by which of the spots you put the ones. In other words, how many ways are there to select six things from a group of ten? This is an important quantity in combinatorics. 
For the second, as a warmup, one might try the easier problem of counting the numbers with exactly two zeros rather than at least. You can break this up into two steps. First is where you put the zeros, and the second is what you put in the other two spots. Using similar reasoning to the first problem (or just counting by brute force), there are six different possibilities for the placement of the two zeros. And then you can assign anything $1-9$ to the other two, yielding $81$ possibilities. So that's $6\times 81 = 486$ possibilities. Now if you repeat the same exercise for exactly three zeros and for four zeros (both of which are easier) and add them, you're done. Alternatively, contra my comment, you could do this by counting those with zero zeros and one zero and subtracting from the total... this way is probably a little easier and it might be a good check to see that they come out the same.
The third one can be done the same way as the second, and will probably be easier by counting the ones that don't obey the criterion as well. (The ones that don't satisfy have either no 1's and no 0's, one 1 and no 0's, one 0 and no 1's, or one 0 and one 1.
