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Matrix mania!

How do you find the number of possible matrices given that N number of elements(different elements) can be used to make a matrix of any order? For example: with 1 element you can make it only 1 matrix of order 1x1, with 2 elements (0,1) we can make 4 matrices of order 1x2, 2x1 and vice versa by interchanging the positions of the two elements respectively, etc. Do reply at thr earliest. I tried making a formula but it doesn't seem to work for 1 element.

Note by Toshali Mohapatra
6 months ago

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\(\begin{align}\text{If }\hspace{2mm}N&=\prod(P_i)^{\alpha_{i}}\hspace{7mm}\color{blue}\text{where, }P_i \text{ denote the distinct prime factors of } N\\\\ \text{Then, } &\left(\prod(\alpha_{i}+1)\right)N!\hspace{4mm} \text{ will be your solution.}\\\\ \text{basically,}&\text{ it is equal to N! times the number of factors of the given number.}\end{align}\)

Anirudh Sreekumar - 6 months ago

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Gr8! Thanks a lot! How did u come up with such a solution?

Toshali Mohapatra - 6 months ago

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The number of possible orders for the matrix is equal to the number of divisors of \(N\) as for each divisor \( D\),

\(N=D \times { \dfrac{N}{D}}\)

and for any matrix of a given order the elements can be swapped in N! ways

Anirudh Sreekumar - 6 months ago

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