Find the number of partitions of the set $\{1, 2, \dots , 25\}$ into $5$ equal parts, where no part is equal to either $\{1, \dots , 5\}$, $\{6, \dots , 10\}$, $\{11, \dots , 15\}$, $\{16, \dots , 20\}$, or $\{21, \dots , 25\}$.
So far I have the following:
$\frac{25!}{(5!)^55!}$ , which is the total possible number of arrangements of a 25 string, divided by $5!^5$ for the double counting of equivalent sets, and one more $5!$ for the counting of permutations of the sets themselves.
$\binom{5}{1} \frac{20!}{(5!)^44!}$, which is the number of ways of fixing one unallowed subset, and counting all the remaining arrangements in a fashion similar to above.
$\binom{5}{2} \frac{15!}{(5!)^33!}$
$\binom{5}{3} \frac{10!}{(5!)^22!}$
$1$, which is the number of ways of fixing four disallowed subsets, which automatically fixes the remaining subset, as the remaining 5 elements inadvertently only allow for 1 more combination.
And from inclusion exclusion, adding these together I get as an answer
$$\frac{25!}{(5!)^55!} - \binom{5}{1} \frac{20!}{(5!)^44!} + \binom{5}{2} \frac{15!}{(5!)^33!} - \binom{5}{3} \frac{10!}{(5!)^22!} + 1$$
Does this logic seem correct? I would appreciate more insight into this problem, as the question I am working with has no provided answer.
There is not much difficulty gained in answering a slightly more general problem than this question, which is the following:
Let $n$ be a positive integer. How many partitions are there of the set $\{1, 2, \ldots , n^{2}\}$ into $n$ equal parts, where no part is equal to either $\{1, \ldots , n\}$, $\ldots$ , or $\{(n-1)n+1, \ldots, n^{2}\}$?
Preliminary Results:
The following result is needed:
Let $k$ and $m$ be positive integers. A set of size $km$ can be partitioned into $k$ subsets of size $m$ in
$$\frac{(km)!}{(m!)^{k}\cdot k!}$$
different ways.
The reasoning for this is already provided in the question above. For additional information about it, see this post.
This will be used in the proof in the case where $m=n$.
Another result that is needed is the inclusion-exclusion principle in the following form:
Let $m$ be a positive integer. Let $C_{1}, \ldots , C_{m}$ be finite sets. Then
$$\Bigg| \bigcup_{i=1}^{m}C_{i} \Bigg| = \sum_{k=1}^{m} (-1)^{k-1} \Bigg( \sum_{\substack{1\leq \ell_{1} < \ldots < \ell_{k} \leq m \\ \ell_{1}, \ldots, \ell_{k} \in \mathbb{Z} }} \Bigg| \bigcup_{i=1}^{k}C_{\ell_{i}} \Bigg| \Bigg).$$
This will be used in the proof in the case where $m=n$ and where $C_{i}$ is the set of partitions of $\{1, \ldots , n^{2}\}$ containing $\{(n-i-1)n+1, \ldots , (n-i)n\}$.
The proof is presented below and is split into three steps.
Step 1:
Throughout the proof, for each $i\in \{1, \ldots , n\}$, define $C_{i}$ to be the set of partitions of $\{1, \ldots , n^{2}\}$ containing $\{(n-i-1)n+1, \ldots , (n-i)n\}$.
Let $k\in \{1, \ldots ,n\}$ and let $\ell_{1}, \ldots \ell_{k}\in \{1, \ldots n\}$ with $\ell_{i} < \ell_{j}$ whenever both $i<j$ and $i,j\in \{1, \ldots , k\}$. It will be shown that
$$\Bigg| \bigcup_{i=1}^{k}C_{\ell_{i}} \Bigg| = \frac{((n-k)n)!}{(n!)^{n-k}\cdot (n-k)!}.$$
Indeed, in this case, $k$ subsets of size $n$ are already fixed, so consequently $kn$ elements of $\{1, \ldots , n^{2}\}$ are already fixed. The remaining $(n-k)n$ elements can vary to form $n-k$ more subsets of size $n$. By the first preliminary result, there are
$$\frac{((n-k )n)!}{(n!)^{n-k}\cdot (n-k)!}$$
such ways to do this. So Step 1 is completed.
Step 2:
It will be shown that
$$ \Bigg| \bigcup_{i=1}^{n}C_{i} \Bigg| = \sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\frac{((n-k)n)!}{(n!)^{n-k}\cdot (n-k)!}.$$
Once this is shown, the desired result will be obtained from this equality in Step 3.
Note that for a fixed $k\in \{1, \ldots , n\}$, the set
$$\{ \{\ell_{1}, \ldots , \ell_{k}\} \subseteq \{1, \ldots , n\}: 1\leq \ell_{1} < \ldots < \ell_{k} \leq n \}$$
corresponds to all the possible subsets of size $k$ in $\{1, \ldots , n\}$. There are $\binom{n}{k}$ such subsets. By applying Step 1 combined with the inclusion-exclusion principle,
\begin{align*} \Bigg| \bigcup_{i=1}^{n}C_{i} \Bigg| &= \sum_{k=1}^{n} (-1)^{k-1} \Bigg( \sum_{\substack{1\leq \ell_{1} \leq \ldots \leq \ell_{k} \leq n \\ \ell_{1}, \ldots, \ell_{k} \in \mathbb{Z} }} \Bigg| \bigcup_{i=1}^{k}C_{\ell_{i}} \Bigg| \Bigg) \\ &= \sum_{k=1}^{n} (-1)^{k-1} \Bigg( \sum_{\substack{1\leq \ell_{1} \leq \ldots \leq \ell_{k} \leq n \\ \ell_{1}, \ldots, \ell_{k} \in \mathbb{Z} }} \frac{((n-k)n)!}{(n!)^{n-k}\cdot (n-k)!} \Bigg) \\ &= \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{((n-k)n)!}{(n!)^{n-k}\cdot (n-k)!} \end{align*}
So Step 2 is completed.
Step 3:
From Step 2,
$$ \Bigg| \bigcup_{i=1}^{n}C_{i} \Bigg| = \sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\frac{((n-k)n)!}{(n!)^{n-k}\cdot (n-k)!}.$$
Because the desired number of partitions is equal to the total number of partitions minus the partitions which contain at least one part of the form $\{(n-i-1)+1, \ldots , (n-i)n\}$ for some $i \in \{1, \ldots , n\}$, and by again applying the first preliminary result, it follows that the desired number of partitions is
$$\frac{(n^{2})!}{(n!)^{n}\cdot n!} - \sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\frac{((n-k)n)!}{(n!)^{n-k}\cdot (n-k)!} = \sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{((n-k)n)!}{(n!)^{n-k}\cdot (n-k)!}.$$
So Step 3 is completed and consequently the proof is completed. $\blacksquare$
For the particular case where $n=5$, by substituting it into the formula above, the number of partitions of the set $\{1, 2, \ldots , 25\}$ into $5$ equal parts, where no part is equal to either $\{1, \ldots , 5\}$, $\ldots$ , or $\{21, \ldots , 25\}$, is equal to
$$\frac{25!}{(5!)^{5}\cdot 5!} - \binom{5}{1}\frac{20!}{(5!)^{4}\cdot 4!} + \binom{5}{2} \frac{15!}{(5!)^{3}\cdot 3!} - \binom{5}{3} \frac{10!}{(5!)^{2}\cdot 2!} + \binom{5}{4} \frac{5!}{(5!)^{1}\cdot 1!} - 1.$$
The difference between the end of this expression compared to yours indicates that what you tried to do by cancelling out some of the expressions at the end did not successfully cancel all of the required duplicate terms out.