How many ways can 4 numbers be arranged using +,-, ÷, and ×? How many ways is it possible to arrange 4 of the same number using addition, subtraction, multiplication and division. Not all of the signs must be used in a combination, for example: x+x+x+x is acceptable. Reverse combinations count as one (x+x+x-x and x-x+x+x).
 A: We have four binary operations, not all of them associative, let alone in combinations. Therefore we have to bother about parentheses. There are five ways to put parentheses among four  operands, namely
$$(ab)(cd),\quad\bigl((ab)c\bigr)d,\quad\bigl(a(bc)\bigr)d,\quad a\bigl((bc)d\bigr)\quad a\bigl(b(cd)\bigr)\ .\tag{1}$$
Each time we have three binary operations performed, which we may choose freely from the four basic operations. It follows that there are $5\cdot 4^3=320$ formally different expressions.
If all four variables are put equal to some "generic" value $t$ then some of these $320$ expressions will be undefined since there is a division by $0$ involved, and a lot of them will have equal value on account of the rules of algebra. It is an interesting programming exercise to list and count the "semantically different" expressions in $t$ resulting in this way. With Mathematica this can be done as follows:
Define a function $f$ of three variables implementing the basic operations  as follows:
$$f(1,x,y):=x+y,\quad f(2,x,y):=x-y,\quad  f(3,x,y):=x*y,\quad f(4,x,y):=x/y\ .$$
Then define a function $g$ of four variables implementing the five terms $(1)$ as follows:
$$g(1,i,j,k):={\tt Simplify}[f(i,f(j,t,t),f(k,t,t))],$$
and so on until $g(5,i,j,k)$. Produce the list of $320$ resulting expressions,  eliminate doubles by using ${\tt Union}$, and remove entries ${\tt Indeterminate}$ etc. by hand. The following picture shows the resulting  final list; it has 64 entries. When, e.g.,  $t=7$ one in fact obtains $64$ different values.

A: you now that arranging any $n$ numbers in a row is equal to $n!$ 
and you can use only 3 signs out of 4 at a time. so first choose 1 sign out of four  (two times) and now choose  1 sign out of three and minus the one case when all numbers are same.
thus the total number of possible combinations = $${4\choose1} {4\choose 1}  {3\choose 1}(4!-4)=960$$
A: Im assuming here that the $x$'s represent the same number. Also Im assuming that the signs $\times,\div,+,-$ represent binary operations and no other symbols are used (as parenthesis), so $-x$ as an unary operation is not taken in account.
Ignoring reflection you have three holes and four different signs to place in them, thus the total number of different orderings of signs will be $4\cdot 4\cdot 4$.
Now: a reflected ordering doenst change if (and only if) the first sign is the same that the last one, then there is a total of $4\cdot 4\cdot 1$ orderings of signs that doesnt change when they are reflected.
Then the number of orderings that change when they are reflected is $4^3-4^2$, thus the total number of valid orderings will be $4^3-(4^3-4^2)/2$, that is, we quit the duplicated orderings under the operation of reflection.
A: I assume parentheses are not considered here, and the standard order of operations is used to evaluate the expression. The result is of the form: $$a_4x^4+a_3x^3+a_2x^2+a_1x+a_0+a_{-1}x^{-1}+a_{-2}x^{-2}$$
I also assume two of the expressions OP describes are equivalent if they are equal as rational expressions in $x$.
Think about the pair $(m,d)$ where $m$ counts how many times $\cdot$ was used, and $d$ counts how many times $\div$ was used. $(m,d)$ must be among the ten options: $$(0,0), (1,0), (2,0), (3,0), (0,1), (1,1), (2,1), (0,2), (1,2), (0,3)$$
$(m,d)=(3,0)$ implies the expression is $\color{blue}{x^4}$.
$(m,d)=(0,3)$ implies the expression is $\color{blue}{x^{-2}}$.
$(m,d)=(2,1)$ implies the expression is $\color{blue}{x^2}$.
$(m,d)=(1,2)$ implies the expression is $\color{blue}{1}$.
$(m,d)=(2,0)$ implies the expression is $\color{blue}{x^3+x}$, $\color{blue}{x^3-x}$, $\color{blue}{2x^2}$, $\color{blue}{0}$, or $\color{blue}{x-x^3}$.
$(m,d)=(0,2)$ implies the expression is $\color{blue}{x^{-1}+x}$, $\color{blue}{x^{-1}-x}$, $\color{blue}{2}$, $0$, or $\color{blue}{x-x^{-1}}$.
$(m,d)=(1,1)$ implies the expression is $\color{blue}{2x}$, $0$, $\color{blue}{x^2+1}$, $\color{blue}{x^2-1}$ or $\color{blue}{1-x^2}$.
$(m,d)=(0,0)$ implies the expression only uses $+$ and $-$, so it's of the form $nx$, and would be one of $\color{blue}{4x}$, $2x$, $0$, or $\color{blue}{-2x}$.
$(m,d)=(1,0)$ implies the expression must be one of  $\color{blue}{x^2+2x}$, $x^2$, $\color{blue}{x^2-2x}$, $\color{blue}{2x-x^2}$, $\color{blue}{-x^2}$ or $\color{blue}{-2x-x^2}$.
$(m,d)=(0,1)$ implies the expression must be one of  $\color{blue}{1+2x}$, $1$, $\color{blue}{1-2x}$, $\color{blue}{2x-1}$, $\color{blue}{-1}$ or $\color{blue}{-2x-1}$.
I count $29$ unique expressions (in blue). Of course this approach may come with human error on my part. Comments with corrections are welcome.
