derangements and permutations in cryptography
This is purely a counting problem. We want the number $N(n,m)$ of possible permutations of $n$ things, with the constraint that only $m$ among these things can map to themselves ($0\le m\le n$). The values in the question are $n=11$, $m=5$.

It holds that $N(n,n)=n!$ (that's the number of permutations of $n$ elements).

It holds that $N(n,0)=\text{}!n=\begin{cases}
1                            &amp;\text{if }n=0\\
\lfloor{{n!+1}\over e}\rfloor&amp;\text{if }n&gt;0\\
\end{cases}$
(that's the number of derangements of $n$ elements).

The computation made of $N(11,5)$ as $5!\cdot\text{}!(11-5)=31,800$ in the question is wrong, as noted: this is less than $!11=14,684,570$, when it should be above (but still less than $11!=39,916,800$). Two mistakes have been made:


the $m=5$ things that can be assigned without constraint among $n=11$ can be so in $n!\over(n-m)!$ ways, not $m!$ ways; fixing that, we would get ${11!\over(11-5)!}\cdot\text{}!(11-5)=14,691,600$, which is more credible;
but after first assigning $m=5$ unconstrained things, the assignments for the $n-m=6$ other things depend on how we arranged the unconstrained things (if we assigned one of the unconstrained things to an initially constrained thing, the later becomes unconstrained); therefore, the above $14,691,600$ is significantly too low.




Taking a deep breath, we can see that $N(n,m)$ can be obtained as
$$\begin{cases}
1                                    &amp;\text{if }n=0, m=0\\
0                                    &amp;\text{if }n=1, m=0\\
n\cdot N(n-1,n-1)                    &amp;\text{if }n&gt;0, m=n\\
(n-1)\cdot N(n-1,1)                  &amp;\text{if }n&gt;1, m=0\\
m\cdot N(n-1,m-1)+(n-m)\cdot N(n-1,m)&amp;\text{if }0&lt;m&lt;n\\
\end{cases}$$
Justifications:


when $n=0, m=0$: there's a single method to map nothing; consistency with $0!=\text{}!0=1$.
when $n=1, m=0$: there's no permutation of one thing that does not map that thing to itself; consistency with $\text{}!1=0$.
when $n&gt;0, m=n$: we assign an unconstrained thing to an unconstrained thing (thus in $n$ possible ways), leaving $n-1$ unconstrained things; that's the classic factorial recursion.
when $n&gt;1, m=0$: we assign a constrained thing to a constrained thing other than itself (thus in $n-1$ possible ways), leaving $n-1$ things among which $1$ became unconstrained.
when $0&lt;m&lt;n$: we assign an unconstrained thing, either

to an unconstrained thing (thus in $m$ possible ways), leaving $n-1$ things among which $m-1$ unconstrained;
to a constrained thing (thus in $n-m$ possible ways), leaving $n-1$ things among which $m$ unconstrained including the one that became unconstrained.



I get $N(13,6)=3,597,143,040$, and $N(11,5)\equiv67\pmod{97}$.



If we don't mind that $m=n+1$ creeps in, $N(n,m)$ can also be obtained as
$$\begin{cases}
(n-1)\cdot N(n-1,1)                  &amp;\text{if }n&gt;0, m=0\\
m\cdot N(n-1,m-1)+(n-m)\cdot N(n-1,m)&amp;\text{if }0&lt;m\le n\\
1                                    &amp;\text{otherwise}\\
\end{cases}$$
or as
$$\begin{cases}
n\cdot N(n-1,n-1)                      &amp;\text{if }n&gt;0, m=n\\
m\cdot N(n-1,m)+(n-m-1)\cdot N(n-1,m+1)&amp;\text{if }0\le m&lt;n\\
1                                      &amp;\text{otherwise}\\
\end{cases}$$
Justification of the new relation when $0\le m&lt;n$: we assign a constrained thing, either


to an unconstrained thing (thus in $m$ possible ways), leaving $n-1$ things among which $m$ unconstrained;
to a constrained thing other than itself (thus in $n-m-1$ possible ways), leaving $n-1$ things among which $m+1$ unconstrained including the one that became unconstrained.




Based on intuition confirmed by numerical evidence that $m\mapsto N(n,m)$ is close to exponential between $m=0$ and $m=n$, I assert that
$$N(n,m)\approx n!\cdot e^{m/n-1}$$
This approximation is by excess when $m&gt;0$. The error made is less than $4.0\%$ for $n\ge4$; less than $1.8\%$ for $n\ge8$; less than $1.0\%$ for $n\ge14$.

More rigourously stated:
$$\forall r\in\mathbb R, 0\le r\le1\implies\lim_{n\rightarrow\infty}{n!\cdot e^{r-1}\over N(n,\lfloor r\cdot n\rfloor)}=1$$