$25$ little donkeys stand in a row; the rightmost of them is Eeyore. Winnie-the-Pooh wants to give a balloon of one of the seven colours of the rainbow to each donkey, so that successive donkeys receive balloons of different colours, and so that at least one balloon of each colour is given to some donkey. Eeyore wants to give to each of the $24$ remaining donkeys a pot of one of six colours of the rainbow (except red), so that at least one pot of each colour is given to some donkey (but successive donkeys can receive pots of the same colour). Which of the two friends has more ways to get his plan implemented, and how many times more?

[i]Eeyore is a character in the Winnie-the-Pooh books by A. A. Milne. He is generally depicted as a pessimistic, gloomy, depressed, old grey stuffed donkey, who is a friend of the title character, Winnie-the-Pooh. His name is an onomatopoeic representation of the braying sound made by a normal donkey. Of course, Winnie-the-Pooh is a fictional anthropomorphic bear.[/i]

[i]Proposed by F. Petrov[/i]
1. **Labeling the Colors and Defining Sets:**
   - Label the colors of the rainbow by \(1, 2, 3, 4, 5, 6, 7\), with red being the 7th color.
   - Define \(A_k\) as the set of ways to give balloons of any but the \(k\)-th color to 25 donkeys, ensuring neighboring donkeys receive balloons of different colors.

2. **Applying the Principle of Inclusion/Exclusion (PIE):**
   - For any \(K \subseteq \{1, 2, 3, 4, 5, 6, 7\}\), the size of the intersection is \(\left|\bigcap_{k \in K} A_k\right| = (7 - |K|)(6 - |K|)^{24}\).
   - Using PIE, we calculate \(\left|\bigcup_{k=1}^7 A_k\right|\):
     \[
     \left|\bigcup_{k=1}^7 A_k\right| = \sum_{\ell = 1}^7 (-1)^{\ell-1} \sum_{|K| = \ell} \left|\bigcap_{k \in K} A_k\right| = \sum_{m = 0}^6 (-1)^m \binom{7}{m} m(m-1)^{24}.
     \]

3. **Calculating the Total Number of Ways Without Restrictions:**
   - Denote \(A_0\) as the set of ways to give balloons of any of the 7 colors to 25 donkeys, ensuring neighboring donkeys receive different colors.
   - Clearly, \(|A_0| = 7 \cdot 6^{24}\).

4. **Finding the Number of Ways Using All 7 Colors:**
   - The number of ways using all 7 colors is:
     \[
     |A_0| - \left|\bigcup_{k=1}^7 A_k\right| = 7 \cdot 6^{24} - \sum_{m = 0}^6 (-1)^m \binom{7}{m} m(m-1)^{24}.
     \]

5. **Calculating the Number of Ways to Distribute Pots:**
   - Define \(B_k\) as the set of ways to give pots of any but red and the \(k\)-th color to 24 donkeys.
   - For any \(K \subseteq \{1, 2, 3, 4, 5, 6\}\), the size of the intersection is \(\left|\bigcap_{k \in K} B_k\right| = (6 - |K|)^{24}\).
   - Using PIE, we calculate \(\left|\bigcup_{k=1}^6 B_k\right|\):
     \[
     \left|\bigcup_{k=1}^6 B_k\right| = \sum_{\ell = 1}^6 (-1)^{\ell-1} \sum_{|K| = \ell} \left|\bigcap_{k \in K} B_k\right| = \sum_{m = 0}^5 (-1)^m \binom{6}{m} m^{24}.
     \]

6. **Calculating the Total Number of Ways Without Restrictions:**
   - Denote \(B_0\) as the set of ways to give pots of any of the 6 colors to 24 donkeys.
   - Clearly, \(|B_0| = 6^{24}\).

7. **Finding the Number of Ways Using All 6 Colors:**
   - The number of ways using all 6 colors is:
     \[
     |B_0| - \left|\bigcup_{k=1}^6 B_k\right| = 6^{24} - \sum_{m = 0}^5 (-1)^m \binom{6}{m} m^{24}.
     \]

8. **Comparing the Two Results:**
   - Notice that:
     \[
     \sum_{m = 1}^7 (-1)^{m-1} \binom{7}{m} m(m-1)^{24} = 7 \sum_{m = 0}^6 (-1)^m \binom{6}{m} m^{24}.
     \]
   - Therefore, the number of ways to distribute balloons is 7 times the number of ways to distribute pots.

The final answer is \(\boxed{7}\).