2.2: Forbidden Position Permutations

Suppose we shuffle a deck of cards; what is the probability that no card is in its original location? More generally, how many permutations of $[n]=\{1,2,3,\ldots,n\}$ have none of the integers in their "correct'' locations? That is, 1 is not first, 2 is not second, and so on. Such a permutation is called a derangement of $[n]$.

Let $S$ be the set of all permutations of $[n]$ and $A_i$ be the permutations of $[n]$ in which $i$ is in the correct place. Then we want to know $|\bigcap_{i=1}^n A_i^c|$.

For any $i$, $|A_i|=(n-1)!$: once $i$ is fixed in position $i$, the remaining $n-1$ integers can be placed in any locations.

What about $|A_i\cap A_j|$? If both $i$ and $j$ are in the correct position, the remaining $n-2$ integers can be placed anywhere, so $|A_i\cap A_j|=(n-2)!$.

In the same way, we see that $|A_{i_1}\cap A_{i_2}\cap\cdots\cap A_{i_k}|=(n-k)!$. Thus, by the inclusion-exclusion formula, in the form of Equation 2.1.1,

$$\eqalign{ |\bigcap_{i=1}^n A_i^c|&=|S|+\sum_{k=1}^n (-1)^k{n\choose k}(n-k)!\cr &=n!+\sum_{k=1}^n (-1)^k{n!\over k!(n-k)!}(n-k)!\cr &=n!+\sum_{k=1}^n (-1)^k{n!\over k!}\cr &=n!+n!\sum_{k=1}^n (-1)^k{1\over k!}\cr &=n!\,\Bigl(1+\sum_{k=1}^n (-1)^k{1\over k!}\Bigr)\cr &=n!\,\sum_{k=0}^n (-1)^k{1\over k!}.\cr }\nonumber$$

The last sum should look familiar: $$e^x=\sum_{k=0}^\infty {1\over k!}x^k.\nonumber$$ Substituting $x=-1$ gives $$e^{-1} = \sum_{k=0}^\infty {1\over k!}(-1)^k.\nonumber$$ The probability of getting a derangement by chance is then $${1\over n!}n!\,\sum_{k=0}^n (-1)^k{1\over k!} =\sum_{k=0}^n (-1)^k{1\over k!},\nonumber$$ and when $n$ is bigger than 6, this is quite close to $$e^{-1} \approx 0.3678.\nonumber$$ So in the case of a deck of cards, the probability of a derangement is about 37%.

Let $D_n=n!\,\sum_{k=0}^n (-1)^k{1\over k!}$. These derangement numbers have some interesting properties. The derangements of $[n]$ may be produced as follows: For each $i\in\{2,3,\ldots,n\}$, put $i$ in position 1 and 1 in position $i$. Then permute the numbers $\{2,3,\ldots,i-1,i+1,\ldots n\}$ in all possible ways so that none of these $n-2$ numbers is in the correct place. There are $D_{n-2}$ ways to do this. Then, keeping 1 in position $i$, derange the numbers $\{i,2,3,\ldots,i-1,i+1,\ldots n\}$, with the "correct'' position of $i$ now considered to be position 1. There are $D_{n-1}$ ways to do this. Thus, $D_n=(n-1)(D_{n-1}+D_{n-2})$.

We explore this recurrence relation a bit:

$$\eqalignno{ D_n&=nD_{n-1}-D_{n-1}+(n-1)D_{n-2}&(*)\cr &=nD_{n-1}-(n-2)(D_{n-2}+D_{n-3})+(n-1)D_{n-2}\cr &=nD_{n-1}-(n-2)D_{n-2}-(n-2)D_{n-3}+(n-1)D_{n-2}\cr &=nD_{n-1}+D_{n-2}-(n-2)D_{n-3}&(*)\cr &=nD_{n-1}+(n-3)(D_{n-3}+D_{n-4})-(n-2)D_{n-3}\cr &=nD_{n-1}+(n-3)D_{n-3}+(n-3)D_{n-4}-(n-2)D_{n-3}\cr &=nD_{n-1}-D_{n-3}+(n-3)D_{n-4}&(*)\cr &=nD_{n-1}-(n-4)(D_{n-4}+D_{n-5})+(n-3)D_{n-4}\cr &=nD_{n-1}-(n-4)D_{n-4}-(n-4)D_{n-5}+(n-3)D_{n-4}\cr &=nD_{n-1}+D_{n-4}-(n-4)D_{n-5}.&(*)\cr }\nonumber$$

It appears from the starred lines that the pattern here is that $$D_n=nD_{n-1}+(-1)^kD_{n-k}+(-1)^{k+1}(n-k)D_{n-k-1}.\nonumber$$ If this continues, we should get to $$D_n=nD_{n-1}+(-1)^{n-2}D_{2}+(-1)^{n-1}(2)D_{1}.\nonumber$$ Since $D_2=1$ and $D_1=0$, this would give $$D_n=nD_{n-1}+(-1)^n,\nonumber$$ since $(-1)^n=(-1)^{n-2}$. Indeed this is true, and can be proved by induction. This gives a somewhat simpler recurrence relation, making it quite easy to compute $D_n$.

$$\bullet\quad\bullet\quad\bullet\nonumber$$

There are many similar problems.

Example $\PageIndex{1}$

How many permutations of $[n]$ contain no instance of $i$ followed by $i+1$?

Solution

By a similar use of the inclusion-exclusion formula, it turns out that this is $$Q_n=n!\,\sum_{k=0}^{n-1} (-1)^k{1\over k!}+ (n-1)!\,\sum_{k=1}^{n-1} (-1)^{k-1} {1\over (k-1)!}. \nonumber$$ Note that the limits on the two sums are not identical.