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consider an experiment that consists of determining the type of job - either blue-collar or white collar- and the political affiliation -republicans, democratic or independent - of the 15 members of an adult soccer team. how many outcomes are in the event that at least one of the team members is a blue-collow worker?

Why is the answer 6^15 - 3^15?

why can't it be 3 * 6^14

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up vote 2 down vote accepted

Our experiment proceeds as follows. We line up the team members, say by Social Security number. We now ask each in turn her/his job description and political affiliation, and record the information.

For any person, the information can be recorded as an ordered pair $(x,y)$, where $x$ is one of $B$ or $W$, and $y$ is one of $R$, $D$, or $I$. So there are for any person $6$ possible records. The team outcome is a sequence of such records, of length $15$. Thus for our team there are $6^{15}$ possible outcomes.

How many of these outcomes have at least one blue-collar worker? How many have no blue-collar worker? Then $x=W$, and $y$ can still take on any of $3$ values, for a total of $3^{15}$ possibilities. Thus the number of possible summaries with at least one $B$ is $6^{15}-3^{15}$.

Remark: What does your answer of $(3)(6^{14})$ count? It seems to count the number of ways that the first person asked is a $B$, and all the rest are free to be anything. Certainly in that case there will be at least one $B$. But there can be at least one $B$ in other ways. We could try to do a fancier version of your idea, but then we can end up double-counting, or worse. That can be compensated for by using the Method of Inclusion/Exclusion, but it gets complicated.

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we just have covered the Method of Inclusion/Exclusion in class today, but the teacher said it'd be easier to get the compliment. that's why he did it this way. How would you approach this problem with the Method of Inclusion/Exclusion? Thank you SOO much for your help! –  user133466 Sep 11 '12 at 1:05
    
We count, like you did, the number of ways in which first person is $B$. Add to this the number of ways second is $B$, third is $B$, and so on. That's $15$ terms exactly like yours. But we have double counted situations in which $i$ and $j$ are $B$. Subtract all these. We have subtracted too much, for we have subtracted too many times the situations where $i$, $j$, and $k$ are $B$, for all choices of three people. And so on. Kind of a mess, but ultimately we get an answer. –  André Nicolas Sep 11 '12 at 1:18
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