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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?

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2 Answers 2

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By designating a particular CD from each genre as the CD from that genre, you count each genre from which Carlos selects more than one CD multiple times.

For instance, if the rap selections are A and B, the country selection is C, and the heavy metal selections are D and E, your method counts this choice $2 \cdot 1 \cdot 2 = 4$ times.

$$ \begin{array}{c c c c} \text{rap} & \text{country} & \text{heavy metal} & \text{additional CDs}\\ \hline A & C & D & B, E\\ A & C & E & B, D\\ B & C & D & A, E\\ B & C & E & A, D \end{array} $$

To avoid this problem, you can either use the Inclusion-Exclusion Principle or you can consider how many CDs of each type are selected.

  • $3$ rap, $1$ country, $1$ heavy metal
  • $2$ rap, $2$ country, $1$ heavy metal
  • $2$ rap, $1$ country, $2$ heavy metal
  • $1$ rap, $3$ country, $1$ heavy metal
  • $1$ rap, $2$ country, $2$ heavy metal
  • $1$ rap, $1$ country, $3$ heavy metal
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  • $\begingroup$ I am not an expert, so, if you are still kind, will you please add a note to the answer to explain your thinking? I have the good feeling that you avoid the real math using some trick, but it beats me. Especially, I do not understand the A, B, C, D, E part, including the associated table. I understand the last table, but I am not sure I get its purpose. Forever grateful. Thank you. $\endgroup$
    – virolino
    Commented Feb 14, 2023 at 5:28
  • $\begingroup$ @virolino The table shows the four ways that JJ H.'s method counts the case that rap CDs A and B are selected, the country CD C is selected, and the heavy metal CDs D and E are selected. We only want to count this case once. To fix the count, we can consider the six cases I listed at the bottom of the post. The number of ways we can select $r$ of the $4$ rap CDs, $c$ of the $5$ country CDs, and $h$ of the $3$ heavy metal CDs is $\binom{4}{r}\binom{5}{c}\binom{3}{h}$. $\endgroup$
    – N. F. Taussig
    Commented Feb 14, 2023 at 10:37
  • $\begingroup$ @virolino Therefore, if we go through the cases in the order I have listed them, the number of selecting $5$ CDs from a collection containing $4$ rap CDs, $5$ country CDs, and $3$ heavy metal CDs if at least one CD from each genre is included is $\binom{4}{3}\binom{5}{1}\binom{4}{1} + \binom{4}{2}\binom{5}{2}\binom{3}{1} + \binom{4}{2}\binom{5}{1}\binom{3}{2} + \binom{4}{1}\binom{5}{3}\binom{3}{1} + \binom{4}{1}\binom{5}{2}\binom{3}{2} + \binom{4}{1}\binom{3}{1}\binom{3}{3}$. $\endgroup$
    – N. F. Taussig
    Commented Feb 14, 2023 at 10:40
  • $\begingroup$ Thank you for the explanation. Now it is more clear. $\endgroup$
    – virolino
    Commented Feb 14, 2023 at 10:47
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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:

$$\frac{[x^5]([(1+x)^4-1][(1+x)^5-1][(1+x)^3-1])}{\binom{12}{5}}=\frac{590}{792}=0,7449...$$

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