How does $\binom{n}{k}\binom{n}{n - k}$ differ from $\binom{n}{k}$? On why the #
of choices  of an $n$ set with $k$ kiwis is $\dbinom{n}{k}\dbinom{n}{n - k} = \dbinom{n}{k}^2$, AlexR wrote:

No, picking exactly $k$ kiwis means you discount the $n-k$ remaining kiwis, but you still have to chose $n-k$ figs out of $n$, which is the second factor.

$1.$ I don't understand AlexR's answer. So I add a picture:  
Why isn't the # of ways of choosing an $n$ set with $k$ kiwis $ \dbinom{n}{k}$?
I know that $\dbinom{n}{k} = \dbinom{n}{n - k}$, which signifies: Out of any selection of $n$ fruits, as soon as you select $k$ kiwis, you simultaneously (de)select the $n - k$ figs. 
But doesn't $\dbinom{n}{k}$ signify both the selection of kiwis and deselection of figs? So isn't multiplying $\dbinom{n}{k}$ by $\dbinom{n}{n - k}$ redundant?   
$2.$ Moreover, how would you determine whether to use $\dbinom{n}{k}\dbinom{n}{n - k}$ or $\dbinom{n}{k}$ ?
 A: This post is a conversion of my comments to an answer.
The confusion lies in the paragraph beginning with "I know that..." The equality of $\binom{n}{k}$ and $\binom{n}{n−k}$ corresponds with the fact that when you select $k$ items from a set of size $n,$ you are, at the same time, deselecting $n−k$ items from that same set. So in your example, when you take $k$ kiwis, you are simultaneously not taking $n−k$ kiwis. As yet, no selection or deselection of figs has taken place. To account for that, you need a second binomial coefficient.
The simple answer to your question two is that you will use a product of two binomial coefficients when you are independently selecting from two different sets; you will use a single binomial coefficient when you are selecting from one set. 
One factor possibly contributing to confusion here is that $n$ plays several distinct roles in this problem: it is the size of the set of kiwis; it is also the size of the set of figs; furthermore, it is the size of the set of selected fruits. It may help to generalize the problem: let there be $r$ kiwis and $s$ figs; select $n$ of those fruits in such a way that there are $k$ kiwis. In this version, selecting $k$ kiwis deselects $r−k$ kiwis. But selecting $n$ fruits, $k$ of which are kiwis, entails selecting $n−k$ figs (and deselecting $s−n+k$ figs).
A: As Will Orrick points out, $n$ plays several roles in your problem, so perhaps generalizing will make this answer a bit clearer.  I will also change the accompanying word problem because I think of pieces of the same type of fruit as being indistinguishable, which is not intended in this setup.
Suppose you have a group of $m$ men and $n$ women from which you must pick a committee of $r$ people.  Certainly there are $\binom{m+n}{r}$ different possible selections.  On the other hand, we can first count how many committees there are with exactly $i$ men for $i = 0, 1, \ldots, r$.  To choose such a committee we first choose $i$ of the $m$ men, which can be done in $\binom{m}{i}$ ways, and to complete the committee, we choose $r - i$ of the $n$ women, which can be done in $\binom{n}{r-i}$ ways.  These choices are independent, so there are $\binom{m}{i} \binom{n}{r-i}$ ways of choosing a committee of size $r$ containing exactly $i$ men.  Summing over the appropriate range of $i$, we see there are $\sum_{i = 0}^r \binom{m}{i} \binom{n}{r-i}$ different choices for a committee of size $r$, hence
$$
\sum_{i = 0}^r \binom{m}{i} \binom{n}{r-i} = \binom{m+n}{r} \, .
$$
