Krdytkyu

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 $^nP_r$ = $frac{n!}{(n-r)!}$

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}.

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

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)! = (3times 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

5C1 is the number of ways to choose 1 item, in any order, from a list of 5 items, A, B, C, D, E.

Let's start by listing the possibilities.

1. A


2. B


3. C


4. D


5. E

5C1=5 different ways

~~~

5C2 is the number of ways to choose 2 items, in any order, from a list of 5 items, for example, A, B, C, D, E.

Let's start by listing the possible combinations starting with A.

1. AB


2. AC


3. AD


4. AE

*It is not possible to obtain AA as there is only a single A.

Next, B.

1. BA


2. BC


3. BD


4. BE

Since for each letter there are 4 ways of choosing the next letter, a direct method would be to take 5 (number of letters) * 4 (different pairs for each letter) = 20 total ways.

However, the order does not matter, meaning that the pair AB and the pair BA does not make a difference. For example, if you are choosing different ways to match 2 fruits out of a selection of 5 fruits (apple, banana, orange, pear, watermelon), whether you have a banana and orange/orange or banana is not important.

Returning to the example of A, B, C, D and E, there is a repeat in the pairing when order is not considered. For example, in AB and BA, there is a repeat of the combination of A and B. This is the same for each of the 5 letters (ED and DE are two of the same thing and so are CD and DC).

Therefore, the total number of combinations is halved.

20/2=10 total ways

Therefore, 5C2 is as such: (5 letters * 4 pairings for each letter)/2 appearances of each pairing = 10 total pairings, simplified as (5*4)/2

~~~

5C3 is the number of ways to choose 2 items, in any order, from a list of 5 items, for example, A, B, C, D, E. This is slightly more challenging.

Let's start by listing the possible combinations starting with A.

Set 1 - with B as the second letter:
1. ABC
2. ABD
3. ABE

Set 2 - with C as the second letter:
1. ACB
2. ACD
3. ACE

Set 3 - with D as the second letter:
1. ADB
2. ADC
3. ADE

Set 4 - with E as the second letter:
1. AEB
2. AEC
3. AED

*Remember that there are no repeats within each possible combination

Notice that, taking A to be the first letter, there are 4 possible letters to choose as the second letter. For each of these second letters, there are 3 possible letters to choose as the third letter (since 2 letters are already used up!).

In other words, when A is the first letter, there are 4 possible sets, and each set has 3 different combinations. Therefore to find the number of possible combinations of A, you can take 4*3=12.

Now, we can replace A with the other 4 letters to obtain the same number of 3-letter combinations, 12 for EACH alphabet. At last, 5 different letters * 4 different sets for each letter * 3 different combinations in each set = 60 total possible ways.

Keeping in mind that the order does not matter, as illustrated through the fruit analogy, we hence must consider that some combinations are repeated, and should be deducted from the total possible ways, 60.

We can do this by finding the number of possible ways to obtain EACH combination of the 60 total combinations. This tells us how many times each combination appears, so we can use the total number to divide by the number of repeats.

Let's look at ABC:

1. ABC


2. ACB


3. BAC


4. BCA


5. CAB


6. CBA

There are exactly 6 ways to reorder one combination. Take a closer look at the list. In a 3-letter combination, there are 3 different letters to choose as the first. For each of these letters, there are 2 different letters to choose as the second. Of course, there is only one left when two letters are chosen, so there is no choice for the third one. Thus 3 ways of choosing the 1st letter * 2 ways of choosing the 2nd letter for EACH first letter, gives you 6 total ways to reorder the letters A, B and C.

This means that each 3-letter combination is repeated 6 times in the total number of combinations. Therefore, to find the number of combinations, without repeats, we can take 60 divided by 6, giving us 10.

Let's refresh how we arrived at these numbers. 5 different letters * 4 ways of choosing the second letter * 3 ways of choosing the third letter = 60 total combinations. 3 ways to rearrange the first letter of EACH combination * 2 ways to rearrange the first letter of EACH combination = 6 ways to rearrange EACH combination. Finally, we divide the total number of combinations by the 6 repeats of each combination.

In other words, 5C3 is as such: 5*4*3/2*3

~~

Take a closer look at 5C1, 5C2 and 5C3.
5C1= 5
5C2= (5*4)/2 = 10
5C3= (5*4*3)/(2*3) = 10

We can rewrite the values (without changing the result, only the expression) to obtain a pattern:
5C1= 5/1
5C2= (5*4)/(2*1)
5C3= (5*4*3)/(3*2*1)

Linking back to the original question, the formula of n!/[r!*(n-r)!] is but another way to express this pattern. It is shown clearly when the values are substituted into the formula.
5C1= [5!/1!(5-1)!] = (5*4*3*2*1)/1(4*3*2*1) = 5/1 - when you cancel out 4*3*2*1
5C2= [5!/2!(5-2)!] = (5*4*3*2*1)/(2*1)(3*2*1) = (5*4)/(2*1) - when you cancel out 3*2*1
5C3= [5!/3!(5-3)!] = (5*4*3**2*1*)/(3*2*1)(2*1) = (5*4*3)/(3*2*1) - when you cancel out 2*1

You may wish to test this out with more numbers if you have not fully grasped the idea. Have fun!

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 $3times2times1=3!=6$ 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: $$ntimes(n-1)timesldots times2times1$$ to: $$ntimes(n-1)timesldotstimes(n-r+1)times(n-r)$$ 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: $$4times3=frac{4times3times2times1}{2times1}=frac{4!}{2!}=frac{n!}{(n-r)!}=12$$ 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. $$frac{n!}{r!(n-r)!}=frac{4!}{2!(4-2)!}=frac{4!}{4}=6$$