1. ## 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?

2. 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

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

3. 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.

4. Oh yeah, where is my head! That'll teach me to try helping people at 2:00AM.