# Thread: Random order of numbers from 1 to 8 - at least six are in their correct slot.

1. ## Random order of numbers from 1 to 8 - at least six are in their correct slot.

Determine the probability of that if you randomly order the numbers from 1 to 8 that at least 6 of them are in their correct slot.

This has to do with probability and factorials. Steps shown would be helpful.

2. Originally Posted by sapphire
Determine the probability of that if you randomly order the numbers from 1 to 8 that at least 6 of them are in their correct slot.
This has to do with probability and factorials. Steps shown would be helpful.
If one rearranges a string of n different items so that no one item appears in its original position that is called a derangement.
The number of derangements of n items is given by $D(n) = n!\left( {\sum\limits_{K = 0}^n {\frac{{\left( { - 1} \right)^k }}{{k!}}} } \right) \approx \frac{{n!}}{e}$.

So to rearrange $12345678$ so that at least six numbers remain in the correct position can be done
in $\sum\limits_{k = 6}^8 {\binom{8}{k}} D(8-k)$ number of ways.