a problem for the probability of distributed objects 17 books and 17 journals are randomly distributed among 17 boy and 17 girl students such that each student gets one item. Find the probability that at least one boy gets a journal.
 A: Hint: It is $1-$ the probability that all the girls get the journals (and all the boys get the books). 
To calculate this probability we have to count all the ways in which the items can be distributed to the students (possible cases, denominator) and all the ways in which all the books are distributed only to the boys (favorable cases, numerator). 
In order to count the ways assume that we can distinguish between the books and the order matters (note that this assumption is not restrective, since we would arrive to the same result if we assumed that the order does not matter (see 2nd way). The only difference is how many cases we count for the denominator and the numerator). Under that assumption we have that
Possible ways. The 34 items can be distributed with $34!$ ways to the students.
Favorable ways. The 17 books should be distributed to the boys. This can be done (since order matters) in $17!$ ways. For each one of these ways the $17$ journals can be distributed with $17!$ ways to the girls. So in total there are $17!\cdot17!$ ways. 
Therefore the asked probability is equal to $$1-\frac{17!17!}{34!}$$
2nd Way. Assume the books and the journals are distinguishable and the order does not matter. Then you can think the process of distributing them as follows. We have a population of $N=34$ items, from which $k=17$ are of interest (the books) and we take a sample of $n=17$ in order to give them to the boys. What is the probability that in our sample we will choose all the books? That is given by the hypergeometric distribution $$\frac{\dbinom{17}{17}}{\dbinom{34}{17}}=\frac{1}{\frac{34!}{17!17!}}=\frac{17!17!}{34!}$$
Therefore the asked probability is again equal to $$1-\frac{17!17!}{34!}$$
However you can assume that the books and the journals are totally indistinguishable. Then the distribution is fully characterized by the number of books that they get. That is, you imagine the process of distributing them as follows. You choose a number in random from $0$ to $17$. Then you give as many books as the number you choosed to the boys (why not?). The rest is then determined automatically. Obviously for 17 out of the possible 18 numbers you can name the boys get at least one book so the asked probability is now equal to $$\frac{17}{18}$$ I would not go for that though.
A: We find first the probability that no boy gets a journal. There are $\binom{34}{17}$ equally likely ways to choose the $17$ people who will get journals. In exactly $1$ of these ways, these $17$ people will be the girls.
Thus the probability that no boy gets a journal is $\frac{1}{\binom{34}{17}}$, and therefore the required probability is $1-\frac{1}{\binom{34}{17}}$.
