What are the different number of matrices available for the following

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The discussion revolves around calculating the number of distinct matrices that can be formed by replacing specific elements in a 5x5 matrix with zeros. The user is interested in understanding how to approach the problem, particularly when considering unique elements and no initial zeros in the matrix. A suggested method involves treating each column separately to determine the arrangements of zeros and then multiplying the possibilities across all columns. The community emphasizes that while they can provide hints, they do not offer direct solutions to encourage independent problem-solving. The conversation highlights the importance of tracking existing zeros in the matrix for accurate calculations.
vcakula
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Greetings,

I have a matrix of order 5 x 5

I would like to replace the

2 elements in column 1 with 0's
1 elements in column 2 with 0's
4 elements in column 3 with 0's
3 elements in column 4 with 0's
2 elements in column 5 with 0's


What are the different number of matrices available if I do replace.

Hoe can this be solved for a general case if matrix is of m x n
 
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Depends on the original matrix. If the original matrix were

\left(\begin{array}{ccccc} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\end{array}\right)

then there would only be one matrix available, no matter what the replacements are.

So, you must keep track of the number of zeroes already in the matrix. For example, you can say that column i has K_i zeroes.
Now, before I give any hints, what are your thoughts about the problem?? What did you attempt?
 
Lets us assume that there are no 0's in the matrix and all the elements of the matrix are unique i.e there are no repetitive elements in the matrix. So the question is like replacing 2 elements in column 1 by 0's.

I tired to treat each column separately as a line with 5 places and check the number of ways 0's could be arranged.

like the
numbers of ways for arranging 2 0's be arranged in 5 places for column 1 and so on for column 2 till column 5.

then multiplied the possibilities of all the columns.

Is this correct..?
 
vcakula said:
Lets us assume that there are no 0's in the matrix and all the elements of the matrix are unique i.e there are no repetitive elements in the matrix. So the question is like replacing 2 elements in column 1 by 0's.

I tired to treat each column separately as a line with 5 places and check the number of ways 0's could be arranged.

like the
numbers of ways for arranging 2 0's be arranged in 5 places for column 1 and so on for column 2 till column 5.

then multiplied the possibilities of all the columns.

Is this correct..?
Sounds right.
 
haruspex said:
Sounds right.

Can you just solve the above problem and let me know the steps which you have followed so that I can compare my answer with yours
 
vcakula said:
Can you just solve the above problem and let me know the steps which you have followed so that I can compare my answer with yours

No, we don't do that here. You need to solve the problem yourself. We will give hints or check the answer. We don't give out answers.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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