I am doing some old exam questions - and I don't know the answer, can some one calculate the result and show how you did it?
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$\begingroup$ I'd suggest starting with the number of ways to pick the scrum master, then look at the options available after this. Working with the one off positions first makes this much easier. $\endgroup$– William AdamczakFeb 8, 2016 at 19:35
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Suppose that A is the person who cannot be scrum master. Let's temporarily forget about the fact A cannot do it. Then the coders can be picked in $\binom{6}{3}$ ways and the rest of the positions can be filled in $3!$ ways, for a total of $\binom{6}{3}3!$ ways.
But this includes the bad ways in which A is scrum master. How many bad ways are there? There are $\binom{5}{3}2!$.
So the total number of good ways is $\binom{6}{3}3!-\binom{5}{3}2!$.
There are many other approaches, at least one of which is simpler.
answered Feb 8, 2016 at 19:39
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1$\begingroup$ Oh i see now. There is 5 ways to choose the scrum master. Then there is 5 ways to choose the tester, and now there is 4 ways to choose a product owner, thus: 5 * 5 * 4 = 100 ways to create the team. Would this also work? $\endgroup$ Feb 8, 2016 at 19:42
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$\begingroup$ Yes, that is one of the simpler ways I referred to, probably the simplest. $\endgroup$ Feb 8, 2016 at 19:47
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$\begingroup$ Yeah. I see the trick here. It's not difficult to calculate it out. But the difficult part i see is to understand the problem :) $\endgroup$ Feb 8, 2016 at 19:51
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$\begingroup$ Understanding the problem can be difficult, since sometimes wording is ambiguous. But here the wording is fine. $\endgroup$ Feb 8, 2016 at 19:52
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$\begingroup$ In your second paragraph, you meant to say the number of bad choices is $\binom{5}{\color{red}{3}}2!$, as you wrote in your final answer. $\endgroup$ Feb 8, 2016 at 20:48