In fact both these numbers can be computed. Label the colours of the rainbow by $1,2,3,4,5,6,7$, with red being the $7$-th. Denote by $A_k$, $1\leq k\leq 7$, the set of ways to give balloons of any but the $k$-th colour to $25$ donkeys, so that neighbouring donkeys receive balloons of different colour (though maybe not all $6$ of the available colours are being used). Clearly, for any $K\subseteq \{1,2,3,4,5,6,7\}$, we have $\displaystyle \left |\bigcap_{k\in K} A_k\right | = (7-|K|)(6-|K|)^{24}$. By the Principle of Inclusion/Exclusion (PIE) we have \[\left |\bigcup_{k=1}^7 A_k\right | = \sum_{\ell = 1}^7 (-1)^{\ell-1}\sum_{|K| = \ell}\left |\bigcap_{k\in K} A_k\right | = \sum_{m = 0}^6 (-1)^{m}\binom {7} {m} m(m-1)^{24}.\] Now, denote by $A_0$ the set of ways to give balloons of any of the colours of the rainbow to $25$ donkeys, so that neighbouring donkeys receive balloons of different colour (though maybe not all $7$ of the available colours are being used). Clearly $|A_0| = 7\cdot 6^{24}$. The number we are looking for, the number of such ways that use all $7$ colors, is therefore \[\displaystyle |A_0| - \left |\bigcup_{k=1}^7 A_k\right |= \sum_{m = 0}^7 (-1)^{m-1}\binom {7} {m} m(m-1)^{24}.\] We have now to compute the number of ways to distribute pots, i.e. $6$ colours (no red), $24$ donkeys, no restriction for neighbouring donkeys, but all $6$ colours to be used. Denote, following the fashion of above, by $B_k$, $1\leq k\leq 6$, the set of ways to give pots of any but red and the $k$-th colour to $24$ donkeys (though maybe not all $5$ of the available colours are being used). Clearly, for any $K\subseteq \{1,2,3,4,5,6\}$, we have $\displaystyle \left |\bigcap_{k\in K} B_k\right | = (6-|K|)^{24}$. By the Principle of Inclusion/Exclusion (PIE) we have \[\left |\bigcup_{k=1}^6 B_k\right | = \sum_{\ell = 1}^6 (-1)^{\ell-1}\sum_{|K| = \ell}\left |\bigcap_{k\in K} B_k\right | = \sum_{m = 0}^5 (-1)^{m-1}\binom {6} {m} m^{24}.\] Now, denote by $B_0$ the set of ways to give pots of any of the colours of the rainbow but red to $24$ donkeys (though maybe not all $6$ of the available colours are being used). Clearly $|B_0| = 6^{24}$. The number we are looking for, the number of such ways that use all $6$ colors, is therefore \[\displaystyle |B_0| - \left |\bigcup_{k=1}^6 B_k\right |= \sum_{m = 0}^6 (-1)^{m}\binom {6} {m} m^{24}.\] All it remains is to notice that, term by term, \[\sum_{m = 1}^7 (-1)^{m-1}\binom {7} {m} m(m-1)^{24} = 7\sum_{m = 0}^6 (-1)^{m}\binom {6} {m} m^{24},\] so there are $\boxed{7}$ more ways to distribute balloons than pots. Alternative Official Solution. This method does not actually count the ways, but rather establishes a one-to-one correspondence between Eeyore's ways to distribute pots and Winnie's ways to distribute balloons, all the while giving Eeyore a red one. In a nutshell, pots and balloons have the same color, except when a pair of two consecutive pots share a same color, in which case the colour of the second balloon in the pair is red. Now, the total number of ways Winnie can distribute balloons is $\boxed{7}$ times larger, since the balloon given to Eeyore may be of any of the $7$ colours, not just red, and clearly for each of the colours of the balloon given to Eeyore there are a same number of ways to do it.

Another way to solve this is first fix the color of Eeyore,$WLOG$ color $7$.Now,fix the ballons(pots) where the first ballon(pot)(our first pot is Eeyore) occured of color $1$,color $2$,etc color $6$.Now,it is easy to see that the number of ways to give ballons from k-color that occured to $k+1$-color are the same,so we are finished.