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I want to express how many ways you can arrange $k$ things in $n$ places.

$$\prod\limits_{k=k}^n k = k (k+1) (k+2)\cdots(n-1) n$$

Edit (added) { The example from which I started thinking about this was placing 3 different letters(always placing each one, and with no repetition, so after arranging,two spaces are left empty) in 5 places. The equivalent form of what I'm asking for would be $\frac{n!}{(n-k)!}$ I think. }

1) Is this a valid product notation (mainly referring to the $k = k$)? If not, why?
2) Is the expression correct for that purpose?
If not: a) how it should be? b) Does it express other thing?

Alejandro Veltri
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    No, write $$\prod_{j=k}^n j $$ instead. At any rate, it is not clear what you mean by "arrange $k$ things in $n$ places", are the places distinguishable, are the things distinguishable? – Pedro Jul 04 '14 at 02:40
  • @PedroTamaroff I think both things and places would be distinguisable. Arranging balls of different colors in numerated transparent boxes makes both distinguishable right? If the things were indistinguishable I would have to divide that by $ k! $, right?. – Alejandro Veltri Jul 04 '14 at 03:53
  • @rewobs If both things and places are distinguishable, you'd simply have $n^k$ possibilities. This assumes you can also leave a place empty. – Magdiragdag Jul 04 '14 at 04:25
  • @Magdiragdag The example from which I started thinking was about placing 3 distinguishable letters(always placing each one, and with no repetition, so after arranging,two spaces are left empty) in 5 places. The equivalent form of what I'm asking for would be $\frac{n!}{(n-k)!}$ I think. I'm editing the main question with this clarification. – Alejandro Veltri Jul 04 '14 at 05:16

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If you have $k$ distinguishable things and $n$ distinguishable places to put them you have $n$ choices to place the first thing. When this first choice has been made you have $n-1$ choices left to place the second thing, and so on. When it comes to place the $k$'th thing there are $n-(k-1)$ empty places left. All in all there are $$N=n(n-1)(n-2)\ldots\bigl(n-(k-1)\bigr)$$ possibilities. Now it comes to write this number $N$ without using "$\ldots$". Since the letters $n$ and $k$ already have been used as parameters of the problem we have to introduce a new symbol, say: $j$, as "multiplication index". In this way we arrive at $$N=\prod_{j=0}^{k-1} (n-j)\ .$$ Another way of representing this number $N$ would be $$N={n!\over(n-k)!}\ .$$

Christian Blatter
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