To solve the problem of finding the number of flags with three strips in order that can be formed using 2 identical red, 2 identical blue, and 2 identical white strips, we can follow these steps:
### Step 1: Identify Possible Combinations
We need to select 3 strips from the available colors (red, blue, and white). The possible combinations of strips can be:
1. 3 strips of the same color (not possible since we only have 2 of each color).
2. 2 strips of one color and 1 strip of another color.
The valid combinations are:
- 2 Red, 1 Blue (R, R, B)
- 2 Red, 1 White (R, R, W)
- 2 Blue, 1 Red (B, B, R)
- 2 Blue, 1 White (B, B, W)
- 2 White, 1 Red (W, W, R)
- 2 White, 1 Blue (W, W, B)
### Step 2: Calculate Arrangements for Each Combination
For each combination, we need to calculate how many distinct arrangements can be made. The formula for arrangements of n items where there are repetitions is given by:
\[
\text{Arrangements} = \frac{n!}{n_1! \cdot n_2! \cdots n_k!}
\]
where \( n \) is the total number of items, and \( n_1, n_2, \ldots, n_k \) are the counts of each distinct item.
#### Arrangements for Each Combination:
1. **2 Red, 1 Blue (R, R, B)**:
\[
\text{Arrangements} = \frac{3!}{2! \cdot 1!} = \frac{6}{2} = 3
\]
2. **2 Red, 1 White (R, R, W)**:
\[
\text{Arrangements} = \frac{3!}{2! \cdot 1!} = \frac{6}{2} = 3
\]
3. **2 Blue, 1 Red (B, B, R)**:
\[
\text{Arrangements} = \frac{3!}{2! \cdot 1!} = \frac{6}{2} = 3
\]
4. **2 Blue, 1 White (B, B, W)**:
\[
\text{Arrangements} = \frac{3!}{2! \cdot 1!} = \frac{6}{2} = 3
\]
5. **2 White, 1 Red (W, W, R)**:
\[
\text{Arrangements} = \frac{3!}{2! \cdot 1!} = \frac{6}{2} = 3
\]
6. **2 White, 1 Blue (W, W, B)**:
\[
\text{Arrangements} = \frac{3!}{2! \cdot 1!} = \frac{6}{2} = 3
\]
### Step 3: Total Arrangements
Now, we sum the arrangements from all valid combinations:
\[
\text{Total Arrangements} = 3 + 3 + 3 + 3 + 3 + 3 = 18
\]
### Step 4: Conclusion
The total number of flags that can be formed using the given strips is **18**.
