"On the Probability of Sequences in the Genoese Lottery" by Euler. How did he do it? This a problem solved by Leonard Euler. Translated English version is available in Euler archives.[E338]On the Probability of Sequences in the Genoese Lottery Euler.
I have difficulty in understanding Corollary 2 on page 29 he derived out of the blue. Can any one explain how did he arrive at  the formula for the number of cases of all species which belong to a particular  value of k. Here's an attached screenshot.
I only want explanation of the first factor in those formulas
(m-1),(m-1)(m-2)/2....
 A: You have the numbers $1,2,3,\ldots,n$ and you choose $m$ of them.  There are ${n \choose m}$ ways of doing this, and that is the final expression.
The question is related to how many sequences of consecutive numbers are in the draw, and $k$ measures this here. So if the drawn numbers were $\{3,4,5,6,7\}$ that would count as $k=1$ sequence, while if the drawn numbers were $\{2,5,6,8,9\}$ that would that would count as $k=3$ sequences.
There is a sign error and the $m-k-1$ should be $m-k+1$ so the initial product is ${m-1 \choose k-1}{n-m+1 \choose k}$.  I doubt species is a good translation of especes; perhaps types might work better.
Both the ${m-1 \choose k-1}$ and ${n-m+1 \choose k}$ are stars and bars calculations.  You want $k-1$ non-empty separators between the sequences and $k$ sequences between the non-drawn numbers, adjusting for the ends of the original set.
In particular the ${m-1 \choose k-1}$ is the number of ways of splitting the $m$ drawn numbers into $k$ sequences, so for example with $m=5$ and $k=2$ we could have $abcd|e$ or $abc|de$ or $ab|cde$ or $a|bcde$ i.e. ${5-1 \choose 2-1}=4$ ways.
