Choice of at least one tutorial group in this Probability question I have this probability question to solve:
Question: k students can choose(randomly) for themselves which one of the n tutorial groups they want to attend, what is the probability that all tutorial groups have at least one student?
My attempt:
P(at least one student) = 1 - P(no student)
P(at least one student) = 1 - (kC0)/(kCn)

I am not sure whether I have done it correctly. Need some guidance.

 A: Let's use inclusion-exclusion.
First, how many ways can the first tutorial group be empty?  This occurs if we give the $k$ students only $n-1$ choices, which leads to $(n-1)^k$ ways.  Counting each tutorial group, we see that there are $n(n-1)^k$ ways that at least one tutorial group is empty if we double count the number of ways in which two tutorial groups are empty.
Since we don't want to double count, let's subtract this number.  How many ways can the first two tutorial groups be empty?  This occurs if we give the $k$ students only $n-2$ choices, which leads to $(n-2)^k$ ways.  Counting all pairs of tutorial groups, we see that there are $\binom{n}{2}(n-2)^k$ ways that at least two tutorial groups are empty if we double count the number of ways in which three tutorial groups are empty.
Continuing as normal, we will find the number of ways that at least one group is empty, without double counting at all:
$$\sum_{i=1}^{n-1}(-1)^{i-1}\binom{n}{i}(n-i)^k$$
This sum is related to Stirling numbers of the second kind, $S(k,n)$, as follows:
$$\sum_{i=1}^{n-1}(-1)^{i-1}\binom{n}{i}(n-i)^k=n^k-n!S(k,n)$$
Now, since there are $n^k$ possible outcomes, the probability that at least one group is empty is obtained by dividing by $n^k$.  Then, to answer your question, subtract this result from $1$.
$$1-\frac{1}{n^k}(n^k-n!S(k,n))=\boxed{\displaystyle\frac{n!S(k,n)}{n^k}}$$
Notice this formula gives $0$ when $k<n$, as intuitively it should.  For $n=2$, it reduces to $1-2^{1-k}$ as is verifiable by considering the probability of obtaining at least one heads and at least one tails after $k$ tosses of a coin.
A: The probability of a particular group not getting chosen by a particular student is (n-1)/n.
The probability of a particular group not getting chosen by any students is ((n-1)/n)^k.
As this can happen to any group, that turns into n(((n-1)/n)^k) that there's at least one group with no students in it.
So the probability of there being no groups with no students in them (or at least one in each) is 1-n(((n-1)/n)^k)
A: First, find the number of tutorial group choices without restrictions.  Take the $k$ students and $n-1$ "dividers".  These can be arranges in a row in $N_{total}$ ways:$$N_{total}= \frac{(k+n-1)!}{k!(n-1)!}$$  Everyone up to the first divider is in the first group, between the first and the second dividers in the second group, and so on.  Clearly, dividers can come together, giving groups with no members.
Now for the restriction.  Starting over, temporarily take $n$ students out of the original $k$ students, and proceed as before.  Take the $k-n$ students and $n-1$ dividers and arrange in a row in $N_{restricted}$ ways:$$N_{restricted}= \frac{(k-1)!}{(k-n)!(n-1)!}$$  Now put the $n$ removed students back, one to each tutorial group.  All groups now have at least one occupant.
The probability that each tutorial group will be occupied by at least one student is :$$P=\frac{N_{restricted}}{N_{total}}$$
