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.

#Algebra

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

$\begin{aligned}\text{If }\hspace{2mm}N&=\prod(P_i)^{\alpha_{i}}\hspace{7mm}\color{#3D99F6}\text{where, }P_i \text{ denote the distinct prime factors of } N\\\\ \text{Then, } &\left(\prod(\alpha_{i}+1)\right)N!\hspace{4mm} \text{ will be your solution.}\\\\ \text{basically,}&\text{ it is equal to N! times the number of factors of the given number.}\end{aligned}$

Anirudh Sreekumar - 4 years, 1 month ago

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

toshali mohapatra - 4 years, 1 month ago

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 - 4 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$