How to use simple examples to explain $^nC_r$ and $^nP_r$. What I mean is not how to use $^nC_r$, $^nP_r$. 
I want examples to explain why $^nC_r$ = $\frac{n!}{r!(n-r)!}$ and  $^nP_r$ = $\frac{n!}{(n-r)!}$
 A: Suppose we want to list all the possible ordered pairs of distinct elements from the set {1, 2, 3, 4, 5}.  One way to do this is first to list all the possible permutations of the set {1, 2, 3, 4, 5}:


*

*12345

*12354

*12435

*12453

*$\vdots$

*54321


and then throw away all but the first two entries in each:


*

*12

*12

*12

*12

*$\vdots$

*54


Of course this produces each pair more than once.  How many times?  Well, how many are there starting with 12?


*

*12345

*12354

*12435

*12453

*12534

*12543


There are six.  In particular, there's one for each permutation of the set {3, 4, 5}.  Similarly, there are six sequences starting with the pair 42, one for each permutation of the set {1,3,5}.  And so on.
So, in all, there are 5! / 3! = 120 / 6 = 20 ordered pairs of two distinct elements of {1, 2, 3, 4, 5}.
Now how many unordered pairs are there?  Well, in the previous count we've counted each pair twice, because there are two ways of ordering it.  So in all there are (5! / 3!) / 2! = 10 unordered pairs of two distinct elements from {1, 2, 3, 4, 5}.
A: I spent some time to think about this
For nPr = n!/(n-r)! 
Assume I have five balls with numbers  1 to 5, I need to select 3 balls in order.
Then the possible outcome would be 5 X 4 X 3 = 5!/2! = 5P3
For nCr = n!/(r!*(n-r)! )
Assume I have five balls with numbers  1 to 5, I need to select 3 balls does not need to be in order.
Since 3 balls can have ABC, ACB, BAC, BCA, CBA, CAB (6 combinations)
Then the possible outcome would be 5 X 4 X 3 / 6 = (5!/2!)/3! = 5C3
A: 
Probability:

How likely something can happen.

Permutation:

Def: The number of possibilities for choosing an ordered set of r objects (a permutation) from a total of n objects.
Definition: $_nP_r(n,r) = n! / (n-r)!$
Assume there are three persons namely A, B and C in the park. But there is only two seats available for them. Then possible ways of people can sit over the seat using permutaion is
{AB, BA, AC, CA, BC, CB} = 6 ways persons can sit on that seats.

$ _nP_r(n,r) = n!/(n-r)! = _3P_2 = 3!/(3-2)! = (3\times 2)/1 = 6$ ways

Combination:

Def: The number of different, unordered combinations of $r$ objects from a set of $n$ objects.
Definition: $_nC_r(n,r) = _nP_r(n,r) / r!  = n!/r!(n-r)!$
Where combination says AB and BA are same since order is doesn't matters.
So {AB, AC, BC} = 3 (combination) ways persons can sit on that seats

$_nC_r(n,r) = n!/r!(n-r)! = _3C_2 = 3!/2!(3-2)! = (3*2)/2 = 3$ ways
A: Let's start with $n!$, this is all the ways to uniquely arrange $n$ elements. For example we can arrange three letters (A, B and C) as ABC, ACB, BAC, BCA, CAB and CBA which is $3!=6$ combinations.
It may help to think of the letters as marked balls in a bag, when you come to pick the first one you have a choice of $3$, then for the second you have a choice of $2$ and for the final pick you can only chose the remaining $1$. This means there are $3\times2\times1=3!=6$ ways you could take the balls out of the bag so $6$ permutations.
$(n-r)!$ can be thought of as a modifier which means you only take $r$ balls out of the bag, not all of them. It changes the equation from: $$n\times(n-1)\times\ldots \times2\times1$$ to: $$n\times(n-1)\times\ldots\times(n-r+1)\times(n-r)$$ If we say we have $4$ balls in a bag and we want to take $2$ of them then for the first choice there is a pick of $4$ balls and for the second a pick of $3$. This means there are: $$4\times3=\frac{4\times3\times2\times1}{2\times1}=\frac{4!}{2!}=\frac{n!}{(n-r)!}=12$$ permutations of taking $2$ elements out of a pool of $4$.
So now we have our permutation equation which cares about the order of elements. The $r!$ can be thought of as a modifier to "unscramble" the permutations and make them into combinations which don't care about the order. As there are $r!$ ways to arrange $r$ elements, dividing by $r!$ "unscrambles" the combinations.
If we think back to our example of taking $2$ balls from a bag of $4$ then we have the  $12$ possibilities: AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, DC. Now if we order these and remove any duplicates we get the $6$ possibilities: AB, AC, AD, BC, BD, CD. $$\frac{n!}{r!(n-r)!}=\frac{4!}{2!(4-2)!}=\frac{4!}{4}=6$$
