Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
Math Topics
»
discretemath
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
How many (0,1)matrices F satisfy E≤F?
Replies:
4
Last Post:
Aug 16, 2013 6:52 PM



Ben Brink
Posts:
201
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



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.



