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$N$ tutors are to be assigned to $s$ students with any student having at most one tutor and similarly any tutor having at most one student. If any tutor is assigned randomly then how can we find the probability that a particular tutor is assigned to a student.

My attempt:

I know that if $s\geq N$ then my answer is one. Further I know that assigning $s$ things from a total of $N\geq s$ without repetition can have $x=\frac{N!}{s!(N-s)!}$. So the probability for my question is $\frac{y}{x}$. The problem is that I do not know how to calculate $y$ with the help of combinations. I will be very thankful for your help. Thanks in advance.

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    $\begingroup$ Assume $N\geq s$. In how many ways can you choose $s$ tutors where you don't care if your specific one is used. In how many ways can you choose $s$ tutors where you guarantee that your specific tutor in mind is used? (To answer the second one, relate it to the question of how many ways you can choose $s-1$ additional tutors from the $N-1$ tutors who are not the special one) $\endgroup$ – JMoravitz Jul 27 '16 at 20:07
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    $\begingroup$ @AustinMohr I interpret it presumably the same way as OP as "what is the probability that alice is working? (it matters not with who)" $\endgroup$ – JMoravitz Jul 27 '16 at 20:08
  • $\begingroup$ @JMoravitz Excellent help thank you. $\endgroup$ – Frank Moses Jul 27 '16 at 20:12
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If $s\geq N$ then there are more students than tutors so, then all tutors are busy and some students are untaught. So the probability that the particular tutor is employed is certainly $1$, in this condition.


If $s\leq N$, some tutors will get the day off.

Thus the denominator, $x$, is the ways to select any $s$ of $N$ tutors to be taught. $$\dfrac{N!}{s!(N-s)!}$$

For the numerator, $y$, we assign count how many ways can our particular tutor be assigned to some student, and then how many ways to select $s-1$ of the remaining the remaining $N-1$ tutors.

Then divide $y$ by $x$, and calculate.


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