For every triple $(a,b,c)$ of non-zero real numbers, form the number \[ \frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}. \] The set of all numbers formed is

$\textbf{(A)}\ {0} \qquad
\textbf{(B)}\ \{-4,0,4\} \qquad
\textbf{(C)}\ \{-4,-2,0,2,4\} \qquad
\textbf{(D)}\ \{-4,-2,2,4\} \qquad
\textbf{(E)}\ \text{none of these}$
To solve this problem, we need to analyze the expression 

\[
\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}
\]

for all possible combinations of signs of \(a\), \(b\), and \(c\). 

1. **Understanding the terms:**
   - \(\frac{a}{|a|}\) is the sign of \(a\). It is \(1\) if \(a\) is positive and \(-1\) if \(a\) is negative.
   - Similarly, \(\frac{b}{|b|}\) is the sign of \(b\), and \(\frac{c}{|c|}\) is the sign of \(c\).
   - \(\frac{abc}{|abc|}\) is the sign of the product \(abc\). It is \(1\) if \(abc\) is positive and \(-1\) if \(abc\) is negative.

2. **Case Analysis:**
   - **Case 1: All positive (\(a > 0\), \(b > 0\), \(c > 0\))**
     \[
     \frac{a}{|a|} = 1, \quad \frac{b}{|b|} = 1, \quad \frac{c}{|c|} = 1, \quad \frac{abc}{|abc|} = 1
     \]
     \[
     1 + 1 + 1 + 1 = 4
     \]

   - **Case 2: Two positive, one negative (\(a > 0\), \(b > 0\), \(c < 0\))**
     \[
     \frac{a}{|a|} = 1, \quad \frac{b}{|b|} = 1, \quad \frac{c}{|c|} = -1, \quad \frac{abc}{|abc|} = -1
     \]
     \[
     1 + 1 - 1 - 1 = 0
     \]
     This result is the same for any permutation of two positive and one negative.

   - **Case 3: One positive, two negative (\(a > 0\), \(b < 0\), \(c < 0\))**
     \[
     \frac{a}{|a|} = 1, \quad \frac{b}{|b|} = -1, \quad \frac{c}{|c|} = -1, \quad \frac{abc}{|abc|} = 1
     \]
     \[
     1 - 1 - 1 + 1 = 0
     \]
     This result is the same for any permutation of one positive and two negative.

   - **Case 4: All negative (\(a < 0\), \(b < 0\), \(c < 0\))**
     \[
     \frac{a}{|a|} = -1, \quad \frac{b}{|b|} = -1, \quad \frac{c}{|c|} = -1, \quad \frac{abc}{|abc|} = -1
     \]
     \[
     -1 - 1 - 1 - 1 = -4
     \]

3. **Conclusion:**
   From the above cases, the possible values of the expression are \(4\), \(0\), and \(-4\). Therefore, the set of all numbers formed is \(\{-4, 0, 4\}\).

The final answer is \( \boxed{ \{-4, 0, 4\} } \)