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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.enter image description here I only want explanation of the first factor in those formulas

(m-1),(m-1)(m-2)/2....

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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.

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  • $\begingroup$ Thank you for explaining but I have few doubts. Euler used an inductive argument to find the second factor so he must have used something like that for first one also ? $\endgroup$
    – kuku
    Commented Oct 12, 2021 at 19:37
  • $\begingroup$ For m=5 and k=2 ,isn't number of split is 2 ? 1{4}+1{1} --> a,a+1,a+2,a+3,b and 1{3}+1{2} --> a,a+1,a+2,b,b+1 page17 of that paper list them $\endgroup$
    – kuku
    Commented Oct 12, 2021 at 19:48
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    $\begingroup$ That earlier section is related to a different question (problem IV in paragraph 14) and different especes. By the time Euler reaches the general problem on page 226 in paragraph 24, he says (1) marque un nombre solitaire and then in paragraph 26 uses that to calculate $k$; if $a$ is $1$ then he counts $\alpha$. The especes in paragraph 26 are the number of sequences including single-number sequences; earlier they were the lengths of sequences. $\endgroup$
    – Henry
    Commented Oct 12, 2021 at 22:01
  • $\begingroup$ English Translation available here :.scholarlycommons.pacific.edu/cgi/… $\endgroup$
    – kuku
    Commented Oct 13, 2021 at 10:30
  • $\begingroup$ Thanks Now got it! But Euler doesn't explicitly say anything about using stars and bars calculations he observes few examples and forms an inductive argument. Am I missing something? $\endgroup$
    – kuku
    Commented Oct 13, 2021 at 10:58

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