Is this answer to this probability question correct?  
A class of $400$.
       • $220$ women.
       • $180$ men.
Divided into study groups of 10 students each.
Q. What is the probability that a certain group will have exactly $4$ men? 

$$ P(x) = \frac{n!}{x! \cdot (n-x)!} \cdot p^x \cdot (1-p)^{(n-x)}$$  

$ n=4 \qquad x=4 \qquad p=0.45 $  
$ P(x) = \frac{10!}{4! \cdot (10-4)!} \cdot 0.45^4 \cdot (1-0.45)^{(10-4)} $  
$ \qquad \,\, =  23.84\%$

This is from the Binomial Distributions topic of a Udemy course I'm taking. That's the given answer. This seems like how to find the probability of a certain group having four men if each person was chosen from $10$ different groups where probability of choosing a man is $0.45$. 
So, is that answer correct? What am I missing?

Lets make a small example: You have a group of 2 men and 3 woman. Now you want divide into a study group of 3 students with 2 women ($Y$). The ways to do that is
$\texttt{wwm, wmw, mww}$
Each group has the same probability. Let us focus on the first group. The probability to choose a woman first is $\frac{3}{2+3}=\frac{3}{5}$
The probability to choose  woman again is $\frac{3-1}{2+3-1}=\frac{2}{4}=\frac{1}{2}$
Here you can see that you have to use a distribution which regard, that the woman are not replaced.
The probability to choose  a man is $\frac{2}{2+3-2}=\frac{2}{3}$
Thus the probability to choose $\texttt{wwm}$ is $\frac{3}{5}\cdot \frac{1}{2}\cdot \frac{2}{3}=\frac1{5}=20\%$. 
Since we have three ways to choose $2$ woman, the probability to choose $2$ woman is $P(Y=2)=60\%$. This is equal to 
$$P(Y=2)=\frac{\binom{3}{2}\cdot \binom{2}{1}}{\binom{5}{3}}$$
