Permutations

A permutation is an ordered arrangement.
By the PERMUTATIONS of the letters abcd we mean all of their possible arrangements:

Using the fundamental concept of counting, If something can be chosen, or can happen, or be done, in m different ways, and, after that has happened, something else can be chosen in n different ways, then the number of ways of choosing both of them is m n.

The first space can be filled by any one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!

Permutations of all

The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n - 1) × (n - 2) ×...× 3 × 2 × 1
Which is also the number of permutations of n different things taken n at a time is n!.

Five different books are on a shelf.  In how many different ways could you arrange them?
Answer.   5! = 1 2 3 4 5 = 120

There are 6! permutations of the 6 letters of the word square.
a)  In how many of them is r the second letter?  _ r _ _ _ _
b)  In how many of them are q and e next to each other?
Solution. a)  Let r be the second letter.  Then there are 5 ways to fill the first spot. After that has happened, there are 4 ways to fill the third, 3 to fill the fourth, and so on.  There are 5! such permutations.
b)  Let q and e be next to each other as qe.  Then we will be permuting the 5 units qe, s, u a, r..  They have 5! permutations.  But q and e could be together as eq.  Therefore, the total number of ways they can be next to each other is 2 5! = 240.

The number of ways of arranging n objects, of which p of one type are alike, q of a second type are alike, r of a third type are alike, etc is:

In how many ways can the letters in the word: STATISTICS be arranged?There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are:

Circular arrangements

The number of ways of arranging n unlike objects in a circle:
Direction matters -> (n – 1)!
Direction doesn't matter -> ½ (n – 1)!

Eight people go to a party. How many different ways can they be seated?
Clockwise and anti-clockwise are different : (8-1)!=7!=5040
Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (8-1)! = 2520

Permutations of less than all

The number of ordered arrangements of r objects taken from n unlike objects is:

For example, imagine putting the letters a, b, c, d into a hat, and then drawing two of them in succession.  We can draw the first in 4 different ways:  either a or b or c or d. After that has happened, there are 3 ways to choose the second.  That is, to each of those 4 ways there correspond 3. Therefore, there are 4X3  or 12  possible ways to choose two letters from four.

ab means that a was chosen first and b second; ba means that b was chosen first and a second; and so on. ”The number of permutations of n different things taken r at a time.” We will symbolize this as :
The lower index 2 indicates the number of factors.  The upper index 4 indicates the first factor. In general,

Combinations

Combinations with repetition

An arrangement of r objects, WITHOUT regard to ORDER and without repetition, selected from n distinct objects is called a combination of n objects taken r at a time. The number of such combinations is denoted by

How many different committees of 4 students can be chosen from a group of 15?
Answer: There are possible combinations of 4 students from a set of 15.
committees.

3 marbles are drawn at random from a bag containing 3 red and 5 white marbles. How many different draws would contain 1 red and 2 white marbles?
This would be a combination problem, because a draw would be a group of marbles without regard to order. It is like grabbing a handful of marbles and looking at them. Ways to pick 1 Red from 3 Red =
Ways to pick 2 White from 5 White =
Total no. of ways = ways X ways = 30

Combinations with repetition

The number of combinations with repetition can be calculated as:

For example, if you have ten types of donuts (n) on a menu to choose from and you want three donuts (k) the number of ways to choose can be calculated as:

There is an easy way to understand this: Let the types of donuts be represented by | and the number of donuts each type was ordered be represented by the number of ∙ Now lets say the first type was ordered 2 times and the 5th type was ordered 1 time making a total of 3 donuts ordered. The corresponding diagram will look like :

That is, 2 donuts (dots) after 1st type (|) and then no dots for 2nd, 3rd, 4th and then 1 donut for 5th type and then again no donuts for the rest. Now the number of dots in our problem will always be 3 and the number of |will always be 10. But the first slash will never move. That is we will never have a situation like this:

because this would not mean anything. So basically we are left with 9 |’s and 3 ∙’s

To generalize, we will have ( n-1) |’s and k ∙’s Number of ways to arrange this will be:

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