The number of solutions, in real numbers $a$, $b$, and $c$, to the system of equations $$a+bc=1,$$$$b+ac=1,$$$$c+ab=1,$$ is

$\text{(A) }3\qquad\text{(B) }4\qquad\text{(C) }5\qquad\text{(D) more than }5\text{, but finitely many}\qquad\text{(E) infinitely many}$
To find the number of real solutions \((a, b, c)\) to the system of equations:
\[
\begin{cases}
a + bc = 1 \\
b + ac = 1 \\
c + ab = 1
\end{cases}
\]
we will analyze the system step by step.

1. **Subtract the first two equations:**
\[
a + bc - (b + ac) = 1 - 1
\]
\[
a - b + bc - ac = 0
\]
\[
(a - b)(1 - c) = 0
\]
This gives us two cases to consider:
\[
\text{Case 1: } c = 1 \quad \text{or} \quad \text{Case 2: } a = b
\]

2. **Case 1: \( c = 1 \)**
\[
a + b \cdot 1 = 1 \implies a + b = 1
\]
\[
b + a \cdot 1 = 1 \implies b + a = 1
\]
\[
1 + ab = 1 \implies ab = 0
\]
Thus, \(a\) and \(b\) must satisfy:
\[
a + b = 1 \quad \text{and} \quad ab = 0
\]
The solutions to these equations are:
\[
(a, b) = (1, 0) \quad \text{or} \quad (a, b) = (0, 1)
\]
Since \(c = 1\), we have the solutions:
\[
(1, 0, 1) \quad \text{and} \quad (0, 1, 1)
\]
Because the equations are symmetric, we can permute these solutions to get:
\[
(1, 0, 1), (0, 1, 1), (1, 1, 0)
\]
Thus, we have 3 solutions from this case.

3. **Case 2: \( a = b \)**
\[
a + ac = 1 \implies a(1 + c) = 1 \implies c = \frac{1 - a}{a}
\]
\[
c + a^2 = 1 \implies \frac{1 - a}{a} + a^2 = 1
\]
Multiply through by \(a\):
\[
1 - a + a^3 = a \implies a^3 - 2a + 1 = 0
\]
We solve the cubic equation \(a^3 - 2a + 1 = 0\). The roots of this equation are:
\[
a = 1, \quad a = -1, \quad a = -1
\]
For each root, we have:
\[
a = b = c
\]
Thus, the solutions are:
\[
(1, 1, 1) \quad \text{and} \quad (-1, -1, -1)
\]
This gives us 2 additional solutions.

Combining the solutions from both cases, we have:
\[
(1, 0, 1), (0, 1, 1), (1, 1, 0), (1, 1, 1), (-1, -1, -1)
\]
Thus, there are 5 solutions in total.

\[
\boxed{5}
\]