What will $_n\!C_r$ and $_n\!P_r$ be when $r=n$ or $r=0$? Suppose I have $n$ items to choose from, and I take all of them, i.e., $r=n$. Then what will be the values of $_n\!C_r$ and $_n\!P_r$?
I think $_n\!C_r$ will be $1$ because there can be only one possible combination while taking all of the items altogether. But I am a little confused about $_n\!P_r$. See, although I can take all items in one shot, will their order matter?
I ask the same question when $r=0$. In this case I think the answer will be $1$ for both $_n\!C_r$ and $_n\!P_r$.
Please correct me if I am wrong?
 A: You are right, $nCn = 1$, because the only way to choose n objects from a set of size $n$ is to take all of them.
However, $nPn = n!$, because you have to take order into account: there are $n$ ways to choose the item in the first position, $n-1$ for the second, and so on (because you can't choose and item twice), so there are $n\cdot (n-1) \cdot (n-2) \cdot \cdots \cdot 1 = n!$ ways to choose all of the elements in order.
Alternatively, if you know the formulas for $nCr$ and $nPr$, you can prove this with them:
$$nCn = {n!\over n!(n-n)!} = {n!\over 0!\cdot n!} = {n!\over n!} = 1$$
$$nPn = {n!\over(n-n)!} = {n!\over 0!} = n!$$
For the second part of your question, you are right that $nC0 = nP0 = 1$, simply because there is only one way to choose no elements from a set, even if you care about the order.
A: You are correct about $nCn$ being equal to $1$, for the precise reason you give. You are also correct about $nPn$ not being $1$ (unless $n=1$), since order does matter. Try to continue with your reasoning to figure this out, you seem to be very close. Also, try using the formulas for $nCr$ and $nPr$, plug $r=n$ into them and see what you get.  
A: Yes it will matter in $nPn$  because(this is choosing and arranging) we have to arrange what was taken out ...
so it is $n!$
