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If I have a 4 digit binary number, how come there are 16 possible "numbers", and similarly, if I have a 4 digit binary number, how come there are 8 possible "numbers?"

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    You just enumerated $2^4=16$ and $2^3=8$. How do you come by the notation of $*$? – 2017-01-10
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    What?, $3*3 \ne 8$ – 2017-01-10
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    What does this have to do with matrices or ternary expansion? Please explain what you are trying to ask -- we will do our best to help you but you have to give us something to work with. – 2017-01-10
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    Explain please my example. – 2017-01-10
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    @Travis That depends on what you mean by $*$. It's not usually used for standard multiplication outside of programming. – 2017-01-10
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    @Arthur, I'm a programmer at heart, that's where I was going with that – 2017-01-10
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    Why will i have 16 variables in the matrix if use number 4 for binary number and 8 variables use number 3 fir binary number? – 2017-01-10

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Still not positive what exactly you're asking, but there are $8$ variations of $3$ digit binary numbers because $2^3 = 8$, and $16$ variations of $4$ digit binary numbers because $2^4 = 16$. To calculate the highest possible number (the amount of variations) in any length binary number, you use $2^n$, $n$ being the length. This applies to base $k$, to calculate the variations or highest number possible in any base, you use $k^n$.

Note that using this formula, your answer will always be in base $10$.

A 5 digit binary number can count to $32$ because $2^5 = 32$

A 5 digit hexadecimal number can count to $1,048,576$ because $16^5 = 1048576$



This works because in base $k$, adding a digit mutliplies the possible numbers by $k$. take Base $2$ as an example, in base $2$, take the number $1001$. If I add 1 digit to the begenning (or end, or anywhere for that matter), there are $2\times$ the number of solutions as before, the number can nowbe $01001$, or $11001$.

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    I think i understood, but not clear, that is, to get the nine options I should use the ternary system. 3^3, not 2^3. – 2017-01-10
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    @Ms.Simpson, there is no length binary number that can only count to $9$, because if $n$ is an integer, $2^n\ne9$, As you stated, $3^3=9$, so it is possible in base three. – 2017-01-10