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

• May 2nd 2010, 12:19 PM
sapphire
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.
• May 2nd 2010, 01:10 PM
Plato
Quote:

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 $\displaystyle D(n) = n!\left( {\sum\limits_{K = 0}^n {\frac{{\left( { - 1} \right)^k }}{{k!}}} } \right) \approx \frac{{n!}}{e}$.

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