What is the probability that Carlos's purchase of 5 CDs includes at least one rap, country, and heavy metal CD out of a total of 12? Full Problem
Carlos has chosen $12$ different CDs he would like to buy: $4$ are rap music, $5$ are country music, and $3$ are heavy metal music. (Carlos has very eclectic tastes in music!) Unfortunately, he has only enough money to afford to buy $5$ of them (they all cost the same price). So he selects $5$ of them at random. What is the probability that his purchase includes at least one CD from each of the three categories?
My Response
First, there are a total of $\dbinom{12}5$ total ways for Carlos to choose, without order, $5$ CDs from $12$ CDs. Then, there are a total of $\dbinom41 \dbinom51 \dbinom31 \dbinom92$ ways for Carlos to choose at least one CD from each category. This simplifies to $\dfrac{30}{11}$, which is obviously not correct. What went wrong in my process?

From another view point:
Because we calculate a probability, each item should be seen as distinct, even if they are in same category. So, let's write their generating functions forms thinking this fact !

*

*Generating function of selecting at least one rap music : $$\binom{4}{1}x^1+\binom{4}{2}x^2+\binom{4}{3}x^3+\binom{4}{4}x^4=(1+x)^4 -1$$


*Generating function of selecting at least one country music : $$\binom{5}{1}x^1+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4+\binom{5}{5}x^5=(1+x)^5 -1$$


*Generating function of selecting at least one heavy metal music : $$\binom{3}{1}x^1+\binom{3}{2}x^2+\binom{3}{3}x^3=(1+x)^3 -1$$
Now , find the coefficient of $x^5$ in the expansion of $$[(1+x)^4-1][(1+x)^5-1][(1+x)^3-1]$$
Calculation:
$$\frac{[x^5]([(1+x)^4-1][(1+x)^5-1][(1+x)^3-1])}{\binom{12}{5}}=\frac{590}{792}=0,7449...$$
