Distributing distinct objects into distinct bins if each bin can hold one object How to distribute $r$ different objects in $n$ different bins? Each bin can hold one object.
The book says that the the answer is:
$$n(n-1)(n-2)(n-3)...(n-r+1)$$ 
I guess this is being done looking at the choices of each object. 
Why can't we look at the problem bin by bin instead like this
$$r(r-1)(r-2)....(r-n+1)$$
I know I am missing a trivial point.
In general how do you select the thing whose point of view is to be taken in account?
 A: It helps to look at extreme cases.
Suppose we have $n$ bins and one object. There are $n$ ways we can match an object with a bin since we have $n$ choices of bin in which we can place the object.  The alternative formula you considered would match all $n$ bins with one object, which would be impractical.
We wish to match objects with bins.  Since only one object can be placed in each bin, we should get zero if there are more objects than bins.  However, if we choose objects rather than bins we will get $0$ whenever there are more bins than objects, which is clearly false for the case of one object and two bins.  Hence, we must choose the bins in which we place the objects.
The formula
\begin{align*}
P(n, k) & = n(n - 1)(n - 2) \ldots (n - k + 1)\\
        & = \frac{n(n - 1)(n - 2) \ldots (n - k + 1)(n - k)!}{k!}\\
        & = \frac{n!}{(n - k)!}
\end{align*}
comes from the observation that we have $n$ bins in which we can place the first object, $n - 1$ remaining bins in which we can place the second object, and so forth until we have $n - (k - 1) = n - k + 1$ remaining bins in which we can place the $k$th object.  Notice that when $k \geq n + 1$, the product reduces to zero, as expected.
A: If we look at the problem bin by bin then the question is something similar like this:


*

*Given $n$ boxes, how many ways are there to put $n$ boxes on $r$ objects if each box can only be placed on one object?

*Otherwise, this question is similar to: Given $n$ different boxes, how many  ways are there to put $r$ objects on each box if each object can only be placed on one box?
We can generalize it like this:
If we need to distribute $r$ objects into $n$ groups, each group hold exactly one object (not each object must be in exactly one group, which is not possible because there may be some leftover objects) then the answer is $n(n-1)(n-2)(n-3)...(n-r+1)$.
A: You can think of two steps: first select $r$ objects in $r!$ ways, then put them into $n$ boxes in ${n\choose r}$ ways. Hence:
$$r!\cdot {n\choose r}=r!\cdot \frac{n!}{r!(n-r)!}=(n-r+1)(n-r+2)\cdots (n-1)n.$$
