What is the probability that there is at least one partygoer who ends up selecting his or her own gift? Suppose that there are 10 people in a party. Each person brings along a gift
for exchange. The gifts are put in a pile and labeled with number 1 ~ 10.
Each person in the party will randomly select one gift by picking slips of
papers with numbers identifying the gifts.
What is the probability that there is at least one partygoer who ends up
selecting his or her own gift?
My solution

Consider the complement of P(At least one)
We want to find P(None)=$\frac{9}{10}*\frac{8}{9}*\frac{7}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5}*\frac{3}{4}*\frac{2}{3}*\frac{1}{2}$

I am stuck. Help please!!!
This is called sampling without replacement. You already know that 

$
P($at least 1 selects his/her own gift$)=1-P($no-one selects his/her gift$)
$

Since the slips are not returned to pile, you need to account for this, hence

$
P($no-one selects his/her gift$) = \frac{9}{10} \cdot \frac{8}{9} \cdots \frac{1}{2}=\frac{1}{10}
$
Hence, 

$
P($at least 1 selects his/her own gift$)=1-\frac{1}{10}=\frac{9}{10}
$

In case sampling was with replacement, i.e. tickets got returned to the pile, this probability converges to (for large number of guests)
$$
\big(1-\frac{1}{n}\big)^n \approx e^{-1}
$$
This is called sampling without replacement. You already know that 

$
P($at least 1 selects his/her own gift$)=1-P($no-one selects his/her gift$)
$

Since the slips are not returned to pile, you need to account for this, hence

$
P($no-one selects his/her gift$) = \frac{9}{10} \cdot \frac{8}{9} \cdots \frac{1}{2}=\frac{1}{10}
$
Hence, 

$
P($at least 1 selects his/her own gift$)=1-\frac{1}{10}=\frac{9}{10}
$

In case sampling was with replacement, i.e. tickets got returned to the pile, this probability converges to (for large number of guests)
$$
\big(1-\frac{1}{n}\big)^n \approx e^{-1}
$$