If 2 out of 6 Machines are Defective, and 3 were selected randomly , What is the Probability Distribution of the number of Defectives?

Even more generally :

If n objects are divided to m alike and k alike , we select x amount randomly , then to find how probable is it that y objects from x are m , using some trial and error I managed to find this complicated formula:

${P}_{X}(y)=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{x}{y}{\textstyle )}{(}_{m}{P}_{y}){(}_{k}{P}_{x-y})}{{}_{n}{P}_{x}}$

But then it turned out that there is an easier formula with Combinations which is actually exactly equal to the previous one

${P}_{X}(y)=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{m}{y}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{k}{x-y}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{n}{x}{\textstyle )}}$

I can't understand how any of them can be shown mathematically or why do they have the same meaning.

Even more generally :

If n objects are divided to m alike and k alike , we select x amount randomly , then to find how probable is it that y objects from x are m , using some trial and error I managed to find this complicated formula:

${P}_{X}(y)=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{x}{y}{\textstyle )}{(}_{m}{P}_{y}){(}_{k}{P}_{x-y})}{{}_{n}{P}_{x}}$

But then it turned out that there is an easier formula with Combinations which is actually exactly equal to the previous one

${P}_{X}(y)=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{m}{y}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{k}{x-y}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{n}{x}{\textstyle )}}$

I can't understand how any of them can be shown mathematically or why do they have the same meaning.