Probability of a flight being overbooked Given an airline that sells $11$ tickets for a flight that only offers $10$ seats and the probability of $10\%$ for each passenger not attending the flight independently of each other, the probability that the flight will be overbooked is $(\frac{9}{10})^{11} \approx 0.31381$.
Now I want to modify the problem to $22$ passengers and $20$ seats. What I first did is $(\frac{9}{20})^{20} \cdot ((\frac{9}{10})^{2}+(\frac{9}{10} \cdot \frac{1}{10})) \approx 0,10941$ which is wrong. The thought behind it was to determine the probability of $20$ passengers attending the flight and at least another passenger attending the flight.
My friend and I think the solution makes more sense like this: 
$(\frac{9}{10})^{20} \cdot ((\frac{9}{10})^{2}+22 (\frac{9}{10} \cdot \frac{1}{10})) \approx 0.3391 $ 
Why do I have to multiply the term by $22$?

In your original problem, the number of passengers arriving is $X \sim \mathsf{Binom}(n=11,p=0.9).$ That is to say, by the binomial PDF formula,
$P(X = k) = {11 \choose k}(0.9)^k(0.1)^{11-k},$ for $k = 0, 1,\dots,11.$
The probability an arriving passenger is denied a seat is 
$P(X > 10) = P(X = 11) = {11 \choose 11}(0.9)^{11}(0.1)^0 = (0.9)^{11},$
as you say.

In your new problem, the number of passengers arriving is
$Y \sim \mathsf{Binom}(n=22,p=0.9).$ That is,
$P(Y = k) = {22 \choose k}(0.9)^k(0.1)^{22-k},$ for $k = 0, 1, \dots, 22.$
The probability an arriving passenger is denied a seat is 
$P(Y > 20) = P(Y=21)+P(Y=22).$ Apply the binomial PDF formula to each of
the two required terms.
As you compute ${22 \choose 21}$, and as per @Muralidharan's hint, you will get the answer to your
question about multiplying by 22.
Using R statistical software, where dbinom is the appropriate binomial PDF,
the computation is as below:
dbinom(21, 22, .9) + dbinom(22, 22, .9)
## 0.3391989

