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Topic: How many (0,1)-matrices F satisfy E≤F?
Replies: 4   Last Post: Aug 16, 2013 6:52 PM

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 Ben Brink Posts: 198 From: Rosenberg, TX Registered: 11/11/06
RE: How many (0,1)-m
atrices F satisfy E≤
F?

Posted: Aug 15, 2013 10:06 PM
 att1.html (1.8 K)

Barry,

That makes sense to me. For E<=F, matrix F can have any of the 0's replaced by 1's (2 choices for each of those entries), thus 2*7 (2 to the 7th power). For G<=E, any of the 1's can be replaced by 0's (2 choices for each of those entries), thus 2*5.

Good work, and notice you can easily calculate the number of matrices F' and G', where E<F' and G'<E. Thanks.

Ben

> Date: Thu, 15 Aug 2013 20:18:12 -0400
> From: discussions@mathforum.org
> To: discretemath@mathforum.org
> Subject: How many (0,1)-matrices F satisfy E?F?
>
> If E=
> [1 0 1 1]
> [0 0 0 1] How many (0,1)-matrices F satisfy E<=F?
> [1 0 0 0] How many (0,1)-matrices G satisfy G<=E?
>
> I am not sure exactly how to define this personally, other than to say that my Text seems to point to this:
> Let E=(e_ij)_(m×n),F=(e_ij )_(m×n),be two m × n (0,1)-matrices. We say that E precedes, or is less than,F,and we write E <= F,if e_ij<=f_ij,for all
> 1 <= i <= m,1 <= j <= n.
>
> So what I think is that I have to matrices which are E&G.
> If I state that there are 5 zeros and 7 ones. this means that E?F= 2^7 because there are 2 matrices and in this instance the zeros are greater than the ones which means 2^7. Opposite is true for G?E which becomes 2^5.
>
> If I am understanding this correctly. I am hoping someone on this forum can help me out with this understanding.

Date Subject Author
8/15/13 Barry Gallegos
8/15/13 Ben Brink
8/15/13 Barry Gallegos
8/15/13 Mark Rickert
8/16/13 Barry Gallegos