Card with 2 numbers, and extraction of 5 numbers from a box with 50 different numbers from 1 to 50. In a game, five numbers between 1 and 50 are extracted from a box. 
I have a card on which are written two numbers between 1 and 50. 
Calculate the probability that both numbers will be drawn from the box. 
The reference solution is: 1 / 123
I have tried to solve without success, as follows:
There are disp (50, 5) = 50*49*48*47*46 = 254251200 ways to extract 5 numbers from the box containing the numbers from 1 to 50.
Leaving out 2 numbers (written on my card), there are disp (48, 3) = 103776 ways to extract 3 numbers from the box.
Leaving out 1 number (written on my card), there are disp (49, 4) = 5085024 ways to extract 4 numbers from the box.
(103776 + 5085024 + 5085024) / 254251200 which is away from the reference solution 1/123.
Hope somebody could help me.
Thank you very much for considering my request.
 A: We give three solutions; the third is the fastest.  We assume the drawing of the $5$ numbers is done without replacement. 
The first way: The standard solution goes as follows. There are $\binom{50}{5}$ ways to choose a five number "hand."  These are all equally likely. 
There are $\binom{48}{3}$ ways to choose $5$ numbers, $2$ of which are the numbers you wrote down.
So the required probability is $\dfrac{\binom{48}{3}}{\binom{50}{5}}$. Calculate. There is a lot of cancellation.
Another way: For an analysis more along the lines of yours, imagine taking the numbers from the box one at a time. Then there are $(50)(49)(48)(47)(46)$ sequences of $5$ numbers, all equally likely.
Now we count the favourable sequences, in which $2$ of the choices match choices you wrote down.
Where are these $2$ numbers?  The location of the smaller one can be chosen in $5$ ways. For each way, the location of the bigger one can be chosen in $4$ ways. 
For each of these locations, $3$ numbers can be placed in the $3$ open slots in $(48)(47)(46)$ ways.
That gives probability $\dfrac{(5)(4)(48)(47)(46)}{(50)(49)(48)(47)(46)}$. Again, there is a lot of cancellation.
Still another way: There are $\binom{5}{2}$ locations where the $2$ good numbers could go. Since all sequences of draws are equally likely, the probability these $2$ locations contain our numbers are the same.
For clarity, let us find the probability that the first and second choices give us our numbers. The probability the first number is one of ours is $\frac{2}{50}$. Given this happened, the probability the second number is one of ours is $\frac{1}{49}$.
This gives probability $\dbinom{5}{2}\cdot \dfrac{2}{50}\cdot\dfrac{1}{49}$.
Remark: The three answers above are the same. Neither is $\frac{1}{123}$, which may be the answer to a different problem. 
A: You can calculate with the converse probability. Thus you have to calculate first the probability, that not both numbers, x and y, will be drawn.
No $x$ and no $y$ are drawn:
$$\frac{{48 \choose 5} \cdot {2 \choose 0}}{50 \choose 5}$$
$x$ and no $y$ are drawn:
$$\frac{{48 \choose 5} \cdot {2 \choose 1}}{50 \choose 5}$$
$y$ and no $x$ are drawn:
$$\frac{{48 \choose 5} \cdot {2 \choose 1}}{50 \choose 5}$$
The difference between 1 and the sum of these three probabilities is very close to 1/123-but not exactly.
