# derangement question

• Dec 7th 2009, 07:57 PM
sbankica
derangement question
A brand new standard deck of 52 playing cards comes with the cards arranged in a particular order. Before using them, the cards are shuffled. What is the probability that exactly 10 of the cards will remain in their original position?
• Dec 7th 2009, 09:21 PM
Bruno J.
Clearly it's equal to the number of possible derangements of the remaining 42 cards, divided by $\displaystyle 52!$. The number of derangements of $\displaystyle n$ objects is the integer closest to $\displaystyle n!/e$, in this case $\displaystyle 51687286544138714389965559095310738479145874768856 1$. So the probability is

$\displaystyle \frac{51687286544138714389965559095310738479145874 7688561}{80658175170943878571660636856403766975289 505440883277824000000000000}$.
• Dec 8th 2009, 02:00 PM
awkward
Quote:

Originally Posted by Bruno J.
Clearly it's equal to the number of possible derangements of the remaining 42 cards, divided by $\displaystyle 52!$. The number of derangements of $\displaystyle n$ objects is the integer closest to $\displaystyle n!/e$, in this case $\displaystyle 51687286544138714389965559095310738479145874768856 1$. So the probability is

$\displaystyle \frac{51687286544138714389965559095310738479145874 7688561}{80658175170943878571660636856403766975289 505440883277824000000000000}$.

Don't forget to take into account the possible choices of the 10 cards that remain in their original position.
• Dec 8th 2009, 02:18 PM
Bruno J.
Oh yeah, where is my head! That'll teach me to try helping people at 2:00AM. (Giggle)