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So I was working on a probability question and then this expression came up.

When I consulted the answers, I struggled to understand exactly how I would get from one expression to the other myself.

enter image description here

Substituting a constant such as

let $n=5$ makes it a bit clearer how they got from one expression to the other.

But is there a simple, yet general explanation of these expressions.

I am hoping the community could give some insight into how I should have approached this problem and similar ones in future.

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Consider the expression $$\prod_{k=1}^{n}\prod_{j=0}^{k-1}(j+1)$$ Can you see what is happening here? Consider each 'block' for lack of a better word, where each block is for some value of $k$. So for the first block due to $k=1$, you have only one term in $\prod_{j=0}^{k-1}(j+1)$, which is $1$. For the second block, $j$ takes two values and so this becomes $1\times2$, and so on; $1$ is repeated $n$ times in the first expression, $2$ is repeated $n-1$ times, and so on. Thus, you can reduce this to a single product as $$\prod_{j=0}^{n}(j+1)^{n-j}$$. Can you now reason similarly for the expression you have?

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Well the first statement says that we are multiplying the expression when $k$ goes from $1$ to $n-1$ AND $j$ goes from $0$ to $k-1$ that is $j<k$.

The second statement says that we are multiplying the expression when $j$ goes from $0$ to $n-1$ AND $k$ goes from $j+1$ to $n$ that is $k>j$. This statement is logically equivalent to the first statement.

The third statement is just a computation of the inner product and as $j$ is not changing for the inner loop, it is being multiplied by itself $n-(j+1)-1$ times.

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Perhaps a picture will help.

Think about the set of points $(j,k)$ in the plane at which you are evaluating the fraction $(40-j)/(52-j)$ (which happens not to depend on $k$). You want to find the product of all the values.

The points (with integer coordinates) will form a triangle. You can think of the product as finding the partial product over the rows first and multiplying, or over the columns first. One way is easier than the other.

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We obtain \begin{align*} P_l(n)&=\prod_{k=1}^n\prod_{j=0}^{k-1}\frac{40-j}{52-j}\\ &=\prod_{\color{blue}{0\leq j<k\leq n}}\frac{40-j}{52-j}\tag{1}\\ &=\prod_{j=0}^{n-1}\prod_{k={j+1}}^n\frac{40-j}{52-j}\tag{2}\\ &=\prod_{j=0}^{n-1}\left(\frac{40-j}{52-j}\right)^{\color{blue}{n-j}}\tag{3} \end{align*}

Comment:

  • In (1) we rewrite the index range to see the relationship somewhat more conveniently.

  • In (2) we change the order of the products and write lower and upper limits corresponding to the index range given in (1).

  • In (3) we note that $\frac{40-j}{52-j}$ does not depend on the index $k$. This means we have a product of $n-j$ times the same factor.

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