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We have a 4 x 4 matrix which can be filled with only 0 and 1. In how many different ways can it be filled? In how many ways can it be filled in symmetry with the first diagonal? The answers are $2^{16}$ and $2^{10}$ but why?

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    what do you exactly mean by "in symmetry with the 1st diagonal" – 2017-02-28
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    @GCab: OP means symmetric with respect to the diagonal from the upper left to lower right. See JMoravitz's answer for further details. – 2017-02-28
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    @BrianTung, ok, thanks, so $A=A^T$ – 2017-03-01

2 Answers 2

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Multiplication principle of counting.

Let our matrix be $\begin{bmatrix}a&b&c&d\\e&f&g&h\\i&j&k&l\\m&n&o&p\end{bmatrix}$

  • Pick the value of $a$ (two options, either $0$ or $1$)
  • Pick the value of $b$ (two options, either $0$ or $1$)
  • $\vdots$
  • Pick the value of $p$ (two options, either $0$ or $1$)

Multiplying the number of options available at each step we get $2^{16}$ different arrangements for the first part of the problem.

For the second part of the problem, since we must have symmetry our matrix looks like

$\begin{bmatrix}a&b&c&d\\b&e&f&g\\c&f&h&i\\d&g&i&j\end{bmatrix}$

  • Pick the value of $a$
  • Pick the value of $b$
  • $\vdots$
  • Pick the value of $j$

Multiplying the number of options available at each step (and making sure we count how many steps there are correctly) we get $2^{10}$ arrangements.

A paraphrasing of the multiplication principle of counting, also sometimes referred to as the rule of product is:

To count the number of outcomes of a scenario, if you can describe every outcome uniquely via a sequence of steps such that the number of options available for a step does not depend on the specific choices made at previous steps (though the available options themselves may vary), the total number of outcomes is the product of the number of options available at each step.

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    I'm unsure why this received a downvote. If the downvoter cares to explain the reason, I may have a chance to improve the post. Otherwise, since the post includes all mathematically correct and factual information, an easy to implement approach, and the correct answers I see nothing to fix. – 2017-02-28
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You can put, repetitions allowed, $\;m\;$ objects in $\;n\;$ holes in $\;m^n\;$ different ways, and thus there are $\;2^{4\cdot4}=2^{16}\;$ differents square matrices of order four with only ones and zeros.

What you seem next to want is to count symmetric matrices, meaning: matrices which remain the same if you switch rows and columns. But such a matrix is completely determined by the elements in its main diagonal and above it, totalling $\;4+3+2+1=10\;$ spots to fill, and thus there $\;2^{10}\;$ such matrices

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    $m$ (*distinct*) objects in $n$ (*distinct*) holes... in $m^n$ different ways? Check that. It should be $n^m$ instead – 2017-02-28