Zuckerberg ,my facebook friend, recently returned from his honeymoon trip to three European cities-Paris, Milan and Zurich, where he had clicked some photos before uploading them on facebook in three different folders naming them on the cities he had visited . The folders Paris, Milan and Zurich have3,4 and 5 photos, respectively. He asks his six year old niece Olivia Bee that if she could download these photos and edit them using the Instagram and put them back but only one photo in each folder. Within no time she edits all the photos in each folder. Within no time she edits all the photos and uploads back quickly one photo in each folder. What's the probability that she uploads at least two photos of the same city and no folder has the photo uploaded back to its original folder, after editing?

Options

A. `1//11`
B. `1//12`
C. `3//11`
D. `1//3`

Answer

To solve the problem, we need to find the probability that Olivia uploads at least two photos of the same city and that no folder has the photo uploaded back to its original folder.

### Step-by-Step Solution:

1. **Identify the Total Number of Photos**:
- Paris: 3 photos
- Milan: 4 photos
- Zurich: 5 photos
- Total photos = 3 + 4 + 5 = 12 photos.

2. **Determine the Total Ways to Choose Photos**:
- Olivia needs to choose one photo from each folder. The number of ways to choose one photo from each folder is:
- From Paris: 3 choices
- From Milan: 4 choices
- From Zurich: 5 choices
- Total ways to choose one photo from each folder = 3 * 4 * 5 = 60 ways.

3. **Determine the Ways to Upload Photos Without Returning to Original Folders**:
- We need to ensure that no photo goes back to its original folder. This is a derangement problem, where we need to find the number of ways to arrange the photos such that none of them is in its original folder.
- The derangement (denoted as !n) of 3 items (since we have 3 folders) is calculated using the formula:
\[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]
- For n = 3:
\[ !3 = 3! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} \right) = 6 \left( 1 - 1 + 0.5 - \frac{1}{6} \right) = 6 \left( 0.5 - \frac{1}{6} \right) = 6 \left( \frac{3}{6} - \frac{1}{6} \right) = 6 \cdot \frac{2}{6} = 2 \]

4. **Calculate the Probability of At Least Two Photos from the Same City**:
- To find the probability that at least two photos come from the same city, we can use the complementary counting method.
- First, calculate the number of ways to select photos such that all come from different cities:
- This is only possible if we select one photo from each city, which is already calculated as 60 ways.
- The total arrangements where no photo goes back to its original folder is 2 (from the derangement calculation).
- Therefore, the number of arrangements where at least two photos come from the same city is:
\[ \text{Total arrangements} - \text{Arrangements with all different cities} = 60 - 2 = 58 \]

5. **Calculate the Probability**:
- The probability that Olivia uploads at least two photos of the same city and no folder has the photo uploaded back to its original folder is:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{58}{60} = \frac{29}{30} \]

### Final Answer:
The probability that she uploads at least two photos of the same city and no folder has the photo uploaded back to its original folder is \( \frac{29}{30} \).