Can number of choices be non-integer? In some lottery, the entry numbers are from $1$ to $80$ inclusive and $22$ numbers are chosen among them. In a ticket someone can choose $10$ numbers and if  his or her $10$ numbers exist among the drawn $22$ numbers he or she wins the jackpot. No repetitive selection is allowed either is draw or ticket. 
So the question is "How many combinations of 10 numbers do you need to play in order to be sure to win?" or in other words "At least how many distinct tickets I have to buy in order to be sure to win?" The answer is (?) : $$\dfrac {C(80,10)}{C(22,10)} = \dfrac {1646492110120}{646646} \simeq 2546203.1933$$ which surprisingly is not integer! If this calculation is correct HOW that's possible and if not where was I wrong? 
 A: What you're computing there is a lower bound for the number of tickets you need to buy. Your computation shows -- if you fix it so the denominator is $\binom{70}{12}$ rather than $\binom{22}{10}$, namely the number of different draws that a single ticket gives win chances for -- that if you buy fewer tickets than the number you get, there will necessarily be possible draws that are not covered by any of your tickets.
What your argument doesn't show is that that this number of tickes -- even if it were an integer -- are enough. It would be enough if you could choose your tickets such that there is no overlap between the set of lucky draws between any two of your tickets. But nobody says that is possible, and in fact for this game it is very impossible. You can't even choose two tickets to play that can't win simultaneously, because those two tickets have at most 20 numbers on them together, and those 20 numbers can all appear in the same draw.
In probability terms, $p={\binom{70}{12}}\big/{\binom{80}{22}}$ is the chance of any given ticket winning. You're trying to reach a winning chance of $1$ by buying $1/p$ different tickets. But in order for that to work, you need to add the probabilities of each of the tickets winning, and that is only valid if the win event for each of your tickets are mutually exclusive. And that is generally impossible here.
A: Always best to look at smaller examples for questions like this.
Let's play a lottery where you pick three numbers from $\{1,2,3,4,5\}$ and the lottery drawing is a pair of numbers from the same set. You win if the pair of numbers are both in you selection.
There are $\binom{5}{2}=10$ pairs of numbers. Each ticket you buy "covers" three pairs of numbers. So you cannot cover all pairs with $3$ tickets. You need at least $10/3$ tickets. Now, $10/3$ is not an integer, it is just a "lower bound" for the number of tickets you need. Since the number of tickets must be an integer, you can say the lower bound is actually $4$.
Now, can you actually cover all possible pairs with $4$ tickets?  Yes, you can:
$$1,2,3\\
1,4,5\\
2,4,5\\
3,4,5$$
Now, ever pair is covered by at least one ticket, so if you buy these four tickets, you are sure of at least one ticket winning.
However, there was nothing guaranteeing that these tickets existed. The numerical calculation I did to get $10/3$ was fairly crude. 
For example, if you pick three numbers from $\{1,2,3,4\}$ and the lottery picks two from $\{1,2,3,4\}$ then you'd get a lower bound of $\frac{\binom{4}{2}}{\binom{3}{2}}=2$, but there is no way to pick two lottery tickets that cover all pairs.
In general, if you are picking $n$ numbers from $\{1,2,\dots,m\}$ and the lottery picks $k\leq n$ numbers, and you win if every number picked by the lottery is in your set, then you need, for $i=1,2,\dots,k$, then, if you can ensure winning with $t$ tickets, you need, for $i=1,\dots,k$:
$$t\binom{n}{i}\geq \binom{m}{i}\left\lceil\frac{(m-i)!(n-k)!}{(n-i)!(m-k)!}\right\rceil$$
That's still only lower bound, not necessarily the greatest lower bound.
For example, with $m=36$, $n=6$, $k=2$, then for $i=2$ we get $t\geq 42$ and for $i=1$ we get $t\geq 42$, but $42$ is known to not be achievable. 
In fact, the question of $(m,n,k)=(n^2,n,k)$ already is an unknown question. Then $t\geq n^2+n$, and we can achieve that value when $n$ is the power of a prime, but it is unknown whether these are the only values of $n$ for which $t=n^2+n$ tickets can ensure a win. 
