2.2: Forbidden Position Permutations
( \newcommand{\kernel}{\mathrm{null}\,}\)
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.
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.