What is the number of permutations possible in a 100 x 100 binary matrix?

2 to the power 10000
2^10000

Are you asking for the number of possible values (2^10,000) or the
number of possible ways to permute the matrix (10,000!)? The number of
permutations is vastly larger than the number of unique values.

> Are you asking for the number of possible values (2^10,000) or the
> number of possible ways to permute the matrix (10,000!)? The number of
> permutations is vastly larger than the number of unique values.

I think I'm asking about the number of possible unique values - if we
were talking about a 1 x 2 binary matrix, the answer would be 4, i.e.
(1 1) (0 1) (1 0) (0 0).

The number is still astronomical 2^10,000 ~ 10^3,000

It is simple. It is a binary 100*100 matrix.
So number of elements are  100*100 = 10000.
for each elemet we can take 2 values either 1 or 0.
total number of ways for 10000 elements = 2*2*... 10000times
                                        = 2^10000

For more genral case,
nxm x-digit matrix, number of diff matrices formed = x^(n*m)

If you are interested in the number of permutations also (original
question, but different than the clarified one), it actually depends
on the given matrix.

The number of permutations of the entries (ie, number of different
arrangements of the entries) depends on the number of entries in the
matrix, in this case 10,000, and the number of 1's (from which we get
the number of 0's, which is equal to 10,000 minus the number of 1's).
As has been discussed, the number of ways to arrange 10,000 numbers is
10,000!, but this assumes all the numbers are distinct. However, since
we are working in binary, some of the entries will be the same since
we only allow 0's or 1's. (For example, the number of permutations of
the matrix (1 0) is two, and the two possible permutations are the
identity, (1 0), and switching the entries (0 1). Further, there is
only one permutation of the matrix (1 1), which is the identity
permutation -- (1 1).)

So if we are given a specific matrix (of 10,000 entries) and the
number of 1's is X, then the number of permutations of that matrix is

     10,000!
------------------
 X! * (10,000-X)!

The 10,000! comes from treating every entry as distinct. However,
since as mentioned above the entries are not actually distinct, we
need the factors in the denominator. The X! accounts for the
overcounting of the number of 1's (the number of ways to move around
all those 1's in each of the "true" permutations), and the (10,000-X)!
accounts for the overcounting of 0's.

If you want to consider a different sized binary matrix, just
substitute the "10,000" in each part of the formula for the number of
entries in the new matrix.

I'm not a mathematician so I may be using the terminology incorrectly,
but I distinguish between the permutation and the result of that
permutation. For this example, the permutation is a function that maps
a 100x100 matrix into another 100x100 matrix with the same matrix
elements, just rearranged. With this definition, the number of
permutations does not depend on the whether the items are distinct,
there are always 10000! different functions of this type. When the
matrix elements are not all distinct (as must be the case for a binary
matrix) the results of these permutations on any specific matrix will
obvioously not all produce unique results.

This is challenging for computer to understand.
Well I think previous commentators got the gist of it. But my opinion
adds possible repetition of digits. I think the matrix is identical
when the elements repeat same number from different elements. Say
there is 1,2,2 and that sequence counts as 1 then 1,2,2 (2's switched
the order) is still sequence 1. What I am saying is that the problem
is the permutation with restricted repetition.
However the problem did not state which numbers exist in the matrix.
Numbers may be integers and range [0,9], or may be they are integers
[0,99]
Assuming that the question deals with the set of integers [a,b] and
let the number of same integers be expressed by colon braket :a: (just
notation).
So my suggested solution to the question is (100*100)!/{ PI_{k=a]^{b}
(:k:)! }, where PI is the infinite product from a (the smallest
integer among the set) to b (the largest integer among the set) The
denominator will cancel out the duplicated integer with respect to the
number of such integers.