Application of simplest combination or permutation clarification

Question

As given in numerous websites, the definition of combination and permutation all one to count different assortments.

My question is that what are the conditions necessary for the simplest (or simple if it is too specific) for a permutation or combination to be applicable? For example- and probably my biggest concern-(Do tell if it is a necessary condition), do the elements in the list need to be indistinguishable to compute one $\binom{n}{k}$? Can elements repeat? Etc...

I ask this because of a question that I have posted (Read the comments of Andre Nicolas's answer for clarification) There are $6$ types of cookies. How many different packs of $3$ cookies can the baker package?

Answer 1

The following information could be useful:

• The twelvefold way

  R.P. Stanley presents in his classic Enumerative combinatorics vol. 1 in section 1.9 the so-called twelvefold way. He considers finite sets $N$ and $X$, with $|N|=n, |X|=x$ and counts the number of different functions $f:N\rightarrow X$ under different situations.

• Functions: $f$ may be arbitrary, injective, surjective giving three different possibilities.

• Sets: Elements of $N,X$ may be either distinguishible or indistinguishible resulting in four different possibilities.

Altogether we can consider $3\cdot 4=12$ different situations:

\begin{array}{llccc} \text{Elements}&\text{Elements}\\ \text{of }N&\text{of }X&\quad\text{Any }f\quad&\quad\text{Injective }f\quad&\quad\text{Surjective } f\quad\\ \hline \text{dist.}&\text{dist.}&x^n\quad&\quad x^{\underline{n}}\quad&x!{n\brace x}\\ \text{indist.}&\text{dist.}&\left(\!\!{x\choose n}\!\!\right)\quad&\quad\binom{x}{n}\quad&\left(\!\!{x\choose n-x}\!\!\right)\quad\\ \text{dist.}&\text{indist.}&\sum_{j=0}^x{n\brace j}\quad&\quad\begin{matrix}1&\text{if }n\leq x\\0&\text{if }n>x\end{matrix}\quad&{n\brace x}\quad\\ \text{indist.}&\text{indist.}&\sum_{j=0}^xp_j(n)\quad&\quad\begin{matrix}1&\text{if }n\leq x\\0&\text{if }n>x\end{matrix}\quad&p_x(n)\quad\\ \end{array}

with

$x^{\underline{n}}=x(x-1)\cdots(x-n+1)$ the falling factorial power,

${n\brace x}$ the Stirling numbers of second kind and

$p_x(n)$ the number of partitions of $n$ into $x$ parts.

I recommend a thorough study of this instructive section.