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Q: number of permutations possible in a 100 x 100 binary matrix? ( No Answer,   8 Comments )
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 Subject: number of permutations possible in a 100 x 100 binary matrix? Category: Science > Math Asked by: baselnovo-ga List Price: \$5.00 Posted: 31 May 2006 12:53 PDT Expires: 30 Jun 2006 12:53 PDT Question ID: 734129
 `What is the number of permutations possible in a 100 x 100 binary matrix?`
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 Subject: Re: number of permutations possible in a 100 x 100 binary matrix? From: philnj-ga on 31 May 2006 13:40 PDT
 ```2 to the power 10000 2^10000```
 Subject: Re: number of permutations possible in a 100 x 100 binary matrix? From: kottekoe-ga on 31 May 2006 15:53 PDT
 ```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.```
 Subject: Re: number of permutations possible in a 100 x 100 binary matrix? From: baselnovo-ga on 31 May 2006 16:45 PDT
 ```> 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).```
 Subject: Re: number of permutations possible in a 100 x 100 binary matrix? From: kottekoe-ga on 31 May 2006 16:50 PDT
 `The number is still astronomical 2^10,000 ~ 10^3,000`
 Subject: Re: number of permutations possible in a 100 x 100 binary matrix? From: coolanakara-ga on 01 Jun 2006 00:04 PDT
 ```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)```
 Subject: Re: number of permutations possible in a 100 x 100 binary matrix? From: youreh-ga on 09 Jun 2006 09:02 PDT
 ```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.```
 Subject: Re: number of permutations possible in a 100 x 100 binary matrix? From: kottekoe-ga on 09 Jun 2006 18:23 PDT
 ```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.```
 Subject: Re: number of permutations possible in a 100 x 100 binary matrix? From: activealexaoki-ga on 10 Jun 2006 12:21 PDT
 ```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.```