# Using a probability argument I got the sum. Can it be done directly?

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• mathman
In summary: I haven't checked. Maybe it can be done directly.As far as I can tell, for each N (or m) it could be done directly.
mathman
TL;DR Summary
Using a probability argument I got the sum. Can it be done directly?
##\sum\limits_{k=0}^{n-m} \frac{\binom{n-m}{k}}{\binom{n}{k}}\frac{m}{n-k}=1##. Can be derived from question. For ##n\ge m##, pick ##m## marbled out of a set size ##n## labeled from ##1## to ##n##, what is probability distribution of minimum of the number labels on the marbles? The terms is the summation are probabilities that minimum label ##=k+1##.

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May we assume ##2m\geq n##?

fresh_42 said:
May we assume ##2m\geq n##?
Error in original post - has been corrected. m can be anything up to n.

We can set ##N:=n-m## and rephrase the statement as
$$\sum_{k=0}^N\prod_{j=1}^k\dfrac{N-j+1}{n-j}=\dfrac{n}{n-N}$$
and attempt an induction on ##N##. It is true for ##N=0,1,2.##
I think that such an induction is the best way to prove it.

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The sum should = 1, why n/m? Also the denominator in the product should be n-j+1.

mathman said:
The sum should = 1, why n/m? Also the denominator in the product should be n-j+1.
No, I distributed ##\dfrac{m}{n}## out of the sum and put it on the other side. My formula is correct (I think). I only skipped a few elementary calculation steps.

Edit: By checking the calculation for ##N=2## I saw that multiplying with the common denominator is probably better than an induction. The induction brought a weighted sum such that the induction hypothesis couldn't be applied.

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At the opposite end, m=n or m=n-1 can be checked directly. Math induction probably won't work. There might be a tricky way to do the sum, but I have no ideas.

There is still a little work to do, but I broke it down to an inductionable statement:
$$\sum_{k=0}^N\prod_{j=1}^k\dfrac{N-j+1}{n-j}=\dfrac{n}{n-N}$$
\begin{align*}
\dfrac{n}{n-N}&= 1+\dfrac{N}{n-1}+\dfrac{N(N-1)}{(n-1)(n-2)}+\ldots+\dfrac{N}{n-1}\cdots \dfrac{1}{n-N}\\
\dfrac{n!}{n-N}&=(n-1)!+N(n-2)!+N(N-1)(n-3)!+\ldots +N!\\
n!&=(n-1)!(n-N)+N(n-N)(n-2)!+\ldots +N!(n-N)
\end{align*}
Now it is in a form where we can perform an induction.
\begin{align*}
N=0\, &: \,(n-1)!(n-0)=n!\\
N=1\, &: \,(n-1)!(n-1)+(n-1)(n-2)!=(n-1)!(n-1+1)=n!\\
N=2\, &: \,(n-1)!(n-2)+2(n-2)(n-2)!+2(n-2)(n-3)!\\
\phantom{N=2\, }&\phantom{\,=}=(n-2)!((n-1)(n-2)+2(n-2)+2)\\
\phantom{N=2\, }&\phantom{\,=}=(n-2)!(n^2-3n+2+2n-4+2)=(n-2)!(n^2-n)=n!
\end{align*}

I see it as a direct calculation. I don't see the induction?

mathman said:
I see it as a direct calculation. I don't see the induction?
I haven't checked. Maybe it can be done directly.

As far as I can tell, for each N (or m) it could be done directly. This is tedious.

## 1. Can you explain the concept of using a probability argument to get a sum?

Using a probability argument to get a sum involves using the principles of probability to calculate the likelihood of a certain sum occurring. This can be done by finding the probabilities of each individual number in the sum and then combining them using mathematical operations.

## 2. How is using a probability argument different from using other mathematical methods to find a sum?

Unlike other mathematical methods, using a probability argument takes into account the likelihood of each individual number in the sum occurring, rather than just their numerical values. This can provide a more accurate and nuanced understanding of the sum.

## 3. Can a probability argument be used to find any type of sum?

Yes, a probability argument can be used to find the sum of any set of numbers, as long as the probabilities of each number can be calculated. This method is particularly useful when dealing with complex or uncertain data.

## 4. Are there any limitations to using a probability argument to find a sum?

One limitation of using a probability argument is that it relies on accurate and reliable data. If the probabilities of the numbers in the sum are not known or are estimated incorrectly, the resulting sum may also be inaccurate.

## 5. Is it possible to directly calculate a sum using a probability argument?

No, a probability argument is used to estimate the likelihood of a sum occurring, rather than directly calculating the sum itself. However, the results of a probability argument can be used to inform and guide the direct calculation of a sum.

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