To solve the problem of finding the number of different resistance values possible using three wires with resistances of 1 ohm, 2 ohm, and 3 ohm, we can explore various combinations of these resistances in series and parallel configurations.
### Step-by-Step Solution:
1. **Understanding Series and Parallel Combinations**:
- In a series combination, the total resistance \( R_s \) is the sum of the individual resistances:
\[
R_s = R_1 + R_2 + R_3
\]
- In a parallel combination, the total resistance \( R_p \) is given by:
\[
\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
2. **Calculating All Possible Combinations**:
- We can use the three resistors in different combinations:
- All three in series
- All three in parallel
- Any two in series and one in parallel
- Any two in parallel and one in series
3. **Calculating Individual Combinations**:
- **All in Series**:
\[
R_{series} = 1 + 2 + 3 = 6 \text{ ohms}
\]
- **All in Parallel**:
\[
\frac{1}{R_{parallel}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = 1 + 0.5 + 0.333 = 1.833 \implies R_{parallel} \approx 0.545 \text{ ohms}
\]
4. **Two in Series, One in Parallel**:
- **(1 ohm and 2 ohm in series, 3 ohm in parallel)**:
\[
R_{s1} = 1 + 2 = 3 \text{ ohms}
\]
\[
\frac{1}{R_{p1}} = \frac{1}{3} + \frac{1}{3} \implies R_{p1} = \frac{3}{2} = 1.5 \text{ ohms}
\]
- **(1 ohm and 3 ohm in series, 2 ohm in parallel)**:
\[
R_{s2} = 1 + 3 = 4 \text{ ohms}
\]
\[
\frac{1}{R_{p2}} = \frac{1}{4} + \frac{1}{2} \implies R_{p2} = \frac{4}{3} \approx 1.333 \text{ ohms}
\]
- **(2 ohm and 3 ohm in series, 1 ohm in parallel)**:
\[
R_{s3} = 2 + 3 = 5 \text{ ohms}
\]
\[
\frac{1}{R_{p3}} = \frac{1}{5} + \frac{1}{2} \implies R_{p3} = \frac{10}{7} \approx 1.428 \text{ ohms}
\]
5. **Two in Parallel, One in Series**:
- **(1 ohm and 2 ohm in parallel, 3 ohm in series)**:
\[
\frac{1}{R_{p4}} = \frac{1}{1} + \frac{1}{2} \implies R_{p4} = \frac{2}{3} \approx 0.667 \text{ ohms}
\]
\[
R_{total} = R_{p4} + 3 = 3.667 \text{ ohms}
\]
- **(1 ohm and 3 ohm in parallel, 2 ohm in series)**:
\[
\frac{1}{R_{p5}} = \frac{1}{1} + \frac{1}{3} \implies R_{p5} = \frac{3}{4} = 0.75 \text{ ohms}
\]
\[
R_{total} = R_{p5} + 2 = 2.75 \text{ ohms}
\]
- **(2 ohm and 3 ohm in parallel, 1 ohm in series)**:
\[
\frac{1}{R_{p6}} = \frac{1}{2} + \frac{1}{3} \implies R_{p6} = \frac{6}{5} = 1.2 \text{ ohms}
\]
\[
R_{total} = R_{p6} + 1 = 2.2 \text{ ohms}
\]
6. **Listing All Unique Resistance Values**:
- From the calculations, we have the following unique resistance values:
- 6 ohms (all in series)
- 0.545 ohms (all in parallel)
- 1.5 ohms (1 & 2 in series, 3 in parallel)
- 1.333 ohms (1 & 3 in series, 2 in parallel)
- 1.428 ohms (2 & 3 in series, 1 in parallel)
- 3.667 ohms (1 & 2 in parallel, 3 in series)
- 2.75 ohms (1 & 3 in parallel, 2 in series)
- 2.2 ohms (2 & 3 in parallel, 1 in series)
7. **Counting Unique Values**:
- The unique resistance values are: 0.545, 1.2, 1.333, 1.428, 1.5, 2.2, 2.75, 3.667, and 6.
- Total unique values = 9.
### Final Answer:
The total number of different resistance values possible using the three wires is **9**.
