Product of $(x-1)(x-2)(x-3)\ldots (x-k+1)$

Question

I was working on a formula, and along the way, I came across this product:$$(x-1)(x-2)(x-3)(x-4)\ldots(x-k+1)\tag1$$ This is just a quick question, but

Question: Is there a shorthand notation for $(1)$? I mean, besides writing all of the terms out.

An example would be something like the sigma notation $\sum$ meaning adding, or $\prod$ meaning the product of the terms.

I feel like there is a notation for $(1)$, but I just don't know it. Any suggestions?

Answer 1

this notion in Mathematics is called a falling factorial (more info from MathWorld). The general notation is $$ (x)_n = x(x-1) \ldots (x-(n-1)), $$ so your expression is $$ (x-1) \times \ldots \times (x-k+1) = (x-1) \times \ldots \times (x-(k-1)) = (x-1)_{k-1} $$

Answer 2

Yes there is. We can write it as $$\prod_{i=1}^{k-1}(x-i).$$

Answer 3

If we multiply the expression by $x$ and divide by $k!$ we have,

$$\frac{x(x-1)....(x-k+1)}{k!}$$

This is precisely ${x \choose k}$, so your product can be rewritten as,

$$k! {x \choose k} \frac{1}{x}$$

After reversing the original operations performed.

If you don't like the $x$ in the denominator here is another solution.

Replace $x$ by $x+1$ in the expression to get,

$$x(x-1).....(x-(k-1)+1)$$

Again we recognize this as,

$$(k-1)! {x+1-1 \choose k-1}$$

Then we replace $x+1$ by $x$ to reverse what we just did,

So the expression is

$$=(k-1)!{x-1 \choose k-1}$$

Answer 4

If falling powers aren't common enough, you could write it fairly economically as $$\frac{(x-1)!}{(x-k)!}$$