45 lines
1.0 KiB
Markdown
45 lines
1.0 KiB
Markdown
#Math #Probability
|
|
|
|
# Problem
|
|
|
|
How many ways are there to arrange a set of $n$ distinct elements such that no element is in it's original position?
|
|
|
|
# Solution
|
|
|
|
The way to arrange the set without consideration for position is:
|
|
|
|
$$
|
|
n!
|
|
$$
|
|
|
|
Now accounting for the values that have one element in it's original position:
|
|
|
|
$$
|
|
n! - {n\choose 1}(n - 1)!
|
|
$$
|
|
|
|
However, we subtracted arrangements with two elements in their original position twice:
|
|
|
|
$$
|
|
n! - {n\choose 1}(n - 1)! + {n \choose 2}(n - 2)!
|
|
$$
|
|
|
|
Now, we readded arrangements with three elements in their original position:
|
|
|
|
$$
|
|
n! - {n\choose 1}(n - 1)! + {n \choose 2}(n - 2)! - {n \choose 3}(n - 3)!
|
|
$$
|
|
|
|
This pattern continues by PIE, giving us:
|
|
|
|
$$
|
|
n! - {n\choose 1}(n - 1)! + {n \choose 2}(n - 2)! - {n \choose 3}(n - 3)! ... + (-1)^n{n \choose n}(n - n)!
|
|
$$
|
|
|
|
Since ${n \choose k}(n - k)! = \frac {n!} {k!}$, we can rewrite as:
|
|
|
|
$$
|
|
\frac {n!} {0!} - \frac {n!} {1!} + \frac {n!} {2!} ... + (-1)^n\frac {n!} {n!} \\
|
|
= \sum _{k = 0}^n (-1)^k \frac {n!} {k!} \\
|
|
= n! \sum _{k = 0}^n \frac {(-1)^k} {k!}
|
|
$$ |