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This is a question given in a previous Statistics test that I am going through (answer provided)

At a gathering, there are $6$ Zimbabweans, $4$ South Africans, $3$ Tanzanians and $4$ Americans. In how many different ways can $6$ people be randomly selected from the gathering in such a way that not all nationalities are represented?
Answer: $12376-6024=6352$

I don't understand where the $6024$ is coming from. Any help would be appreciated. Here is my working:

$n(\text{Not all represented})= n(\text{Total}) - n(\text{All Represented})$
$= (^{17}C_6) - (^6C_1)(^4C_1)(^3C_1)(^4C_1)(^{13}C_2)$
$= 12376 - 22464$
$= -10088$

which is the most obviously wrong answer I have ever obtained. My working in the second term is that from each nation we choose $1$, and after this selection is complete, we choose $2$ more people from the remaining $13$ to make up a group of $6$ such that all nationalities are represented. (Then subtracting this from the total).

Harsh Kumar
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Delonjnaidu
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1 Answers1

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The error in your argument for $n(\text{All Represented})$ is that it counts all such selections multiple times.

Take for example 3 Zimbabweans ($Z_1,Z_2,Z_3$) and one each from South Africa ($S$), Tanzania ($T$) and America ($A$). You count them once for first selecting ($Z_1,S,T,A$) , then choosing ($Z_2,Z_3$) for the remaining 2. But you also count them for first selecting ($Z_2,S,T,A$) , then ($Z_1,Z_3$) and for ($Z_3,S,T,A$) /($Z_1,Z_2$).

For that reason, I cannot calculate $n(\text{All Represented})$ in one simple single formula. You have to look at all the possible ways how the selected 6 could be distributed among the 4 nations. If we use the above abbreviations you get: $$\begin{array}{|c|c|c|c|} Z(6) & S(4) & T(3) & A(4) \\ \hline 3 & 1 & 1 & 1 \\ 1 & 3 & 1 & 1\\ 1 & 1 & 3 & 1\\ 1 & 1 & 1 & 3\\ 2 & 2 & 1 & 1\\ 2 & 1 & 2 & 1\\ 2 & 1 & 1 & 2\\ 1 & 2 & 2 & 1\\ 1 & 2 & 1 & 2\\ 1 & 1 & 2 & 2\\ \end{array} $$

The first row gives you $(^6C_3)(^4C_1)(^3C_1)(^4C_1) = 960 $ possible selections. You can continue to do that for all the other rows and the sum will give you $n(\text{All Represented})$.

Ingix
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