Why in a Binomial distribution there is a binomial coefficient? Scenario:


*

*Supposing an experiment where T aligned bottles are transported.

*Supposing the probability of each bottle being broken during transport is p.


And knowing the bottles can break in any order in the line, 
Question:
why do we consider a binomial coefficient (combination) instead of a permutation, for the probability density function?
$\binom{1000}{n}\times p^n \times q^{T-n}$
One bottle might break, but it is not the same thing being it the first than the 54th.
 A: It's a combination (not a variation) because the order of elements within the "selection" (the broken bottles) doesn't matter.

it is not the same thing being it the first than the 54th.

Of course. But that's taken into acount by the binomial coefficient. Both the combination and the variation considers distinguishable elements. The difference among them is whether the order of the selected elements matter. Here it doesn't matter, so it's a combination. It would matter (and hence be a variation) if, for example, the event "bottles 2 and 10 broke" were different from the event "bottles 10 and 2 broke".
A: Binomial coefficient leterally means: arrang n unique items in any order then choose positins for k of them such that neither the order within these positions, nor positins of other elements matter, i.e. only the numbers of selected positions matter.  Hence in Binom distribution all that matters is the number of successes.
A: You need to get a couple of points clear.


*

*We are dealing with a probability mass function, not a probability density function.

*The binomial distribution gives the (discrete) probability distribution of the number of successes, doesn't consider the order in which they occur, in the sense that suppose bottle $1$ and bottle $49$ break, two bottles have broken, it doesn't distinguish as to which broke first.
A: 
why do we consider a binomial coefficient (combination) instead of a permutation, for the probability density function?

Because we are counting the distinct permutations of $n$ objects which are identical except for type: $k$ broken bottles and $n-k$ unbroken bottles.
$\binom{n}{k}$ counts the ways to select $k$ from $n$ distinct items.   When the distinct "items" being selected are positions, $\binom{n}{k}$ counts the ways to arrange $k$ and $n-k$ objects of two types.   Thus the coefficient counts the permutations of $k$ successes in $n$ trials.
