# Equivalent Capital Pi Notation Expressions

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. 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.

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?

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.

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.

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.

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

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.