Notation about n choose k permutations

Question

I'm stuck on how to write the notation formally and clearly for the set over all the product in the form of $\lambda_{i_1}\lambda_{i_2}\dots\lambda_{i_r}$, where each $i_j$ could be $1,2,\dots,$ or $n$, and $n$ is greater than $r$. Note that something like $\lambda_{1}\lambda_{1}\dots\lambda_{1}\lambda_{1}$ (repeating $\lambda_{1}$ $r$ times) or $\lambda_{1}\lambda_{1}\dots\lambda_{1}\lambda_{2}$ (repeating $\lambda_{1}$ $(r-1)$ times followed by a single $\lambda_2$) should also be included in the set. The multiplication is supposed to be commutative, so the order doesn't matter.

My current attempt: $$ \left\{\prod_{j=1}^{r}\lambda_{i_j}:i_j\in\{1,2,\dots,n\}\right\} $$

It doesn't seem clear enough. Does anyone have a better idea? I really appreciate it so much!

Answer 1

Basic notation:

• $\ \mathbb Z_+\ :=\ \{\,n\in\mathbb Z:\ n\ge 0\,\}$
• $\ \mathcal R$ -- an arbitrary unitary commutative ring, $\ 1\in\mathcal R;$
• $\ V $ -- an arbitrary set (finite or infinite);   it should be interpreted as a set of variables, as a generalization of the OP's set $\ \{\lambda_1\,\ldots\,\lambda_n\}.$

Remark:   Please, do not "correct" my notation (neither above nor below).

OP, your notation is fine and clear to me. It stands for the set of the monomials of degree $r,\ $ in $\ n\ $ variables (a variable doesn't have to appear explicitly in a monomial).

Possibly, the given expression for the number of monomials is not clear to mathematicians. It is indeed awkward but for objective reasons of no fault of OT.

The mentioning of a (commutative) ring is not crucial, it just tells us that the variables are commutative. Thus, we can simply talk about abstract monomials in commuting variables.

One way or another, I'll simply add an equivalent formula, it should be useful in any event. Thus, let $\ \mathcal R\ $ be an arbitrary commutative ring, and we consider elements $\ \lambda_i\ $ to be variables that run over $\ \mathcal R.\ $ OP (@Thinkpad) has defined:

$$ \Lambda_0(n\ r)\ :=\ \left\{\prod_{j=1}^{r}\lambda_{i_j}:i_j\in\{1,2,\dots,n\}\right\} $$

(just name $\ \Lambda_0\ $ is mine). This looks visually simple but actually is not simple conceptually. I'll write something visually not so simple but structural.

Let $$ J(V\ r)\ :=\ \left\{\ j\in\mathbb Z_+^{\ V}: \ \sum j\,=\,r\,\right\} $$ Then the general expression$\ \Lambda\ $ is defined as follows: $$ \Lambda(V\ r)\ :=\ \left\{\,\prod j:\ j\in J(V\ r)\,\right\} $$

This finally is a clean definition of monomials of degree $\ r\ $ (in variable for $\ V$).

We see that:

$$ \Lambda_0(n\ r)\ =\ \Lambda(\{1\,\ldots\,n\}\,\ r) $$

Answer 2

We use the common notation $[n]:=\{1,2,\ldots,n\}$ and consider as index set the cartesian product \begin{align*} [n]^r=\{\left(x_1,x_2,\ldots,x_r\right)\big|x_k\in[n], 1\leq k\leq r\} \end{align*}

Using this notation we can write the set of wanted products as \begin{align*} \color{blue}{\left\{\prod_{k=1}^r\pi_k(x)\Big|x\in[n]^r\right\}} \end{align*}

Here we use the projection operator $\pi_k$ to select the $k$-th component of an $r$-tuple: \begin{align*} \pi_k(x)=\pi_k(x_1,x_2,\ldots,x_r)=x_k \end{align*}

Answer 3

By the commutative property of multiplication (so that $\lambda_2 \cdot \lambda_1 \cdot \lambda_1 = \lambda_1 \cdot \lambda_2 \cdot \lambda_1 = \lambda_1 \cdot \lambda_1 \cdot \lambda_2$) your notation always works.
The only minor edit that I would perform, is to compactly specify something as:
Let $r,n \in \mathbb{Z}^+$, we have the set $$ \left\{\prod_{j=1}^{r}\lambda_{i_j}:i_j\in\{1,2,\dots,n\}, 1 \leq r < n \right\}\qquad $$ $$\qquad =\ \bigcup_{r=1}^{n-1} \left\{\prod_{j=1}^{r}\lambda_{i_j}:i_j\in\{1,2,\dots,n\}\right\} $$

This is fine (IMHO), knowing in advance that $\{\lambda_{i_1} \cdot \lambda_{i_2} \cdots \lambda_{i_{n-1}} \cdot \lambda_{i_n} \}$ ($n$ factors) can never belong to the above mentioned set (since $r<n$ by hypothesis).