Permutation and combinations

Permutation
When a group of objects or people are arranged in a certain order, this arrangement is called a permutation
For permutations, the order is important (for combinations, order is unimportant.
You can find the number of permutations by making use of the fundamental counting principal and multiplying the choices for each category together
Or you can use the following formula:
The number of permutations of n distinct objects taken r at a time is given by:
P(n,r) = n! / (n – r)!
Can’t use if choices may be repeated!

Example
Eight people enter the Best Pic contest. How many ways can blue, red, and green ribbons be awarded?
Since each winner will receive a different ribbon, order is important. You must find the number of permutations of 8 things taken 3 at a time.
p_r^n=n!/((n-r)!) Permutation formula
=8!/((8-3)!) n=8,r=3
=8!/5!=(8×7×6×5!)/5!
=336
The ribbons can be awarded in 336 ways.

Your turn
Example
Ten people are competing in a swim race where 4 ribbons will be given. How many ways can blue, red, green, and yellow ribbons be awarded?
Answer : 5040 ways
Example
From a class of 20 students we need to select 3 for a committee, one to be president, another one to be vice-president and the third one to be secretary.
In this case n = 20, and r = 3. The order is important, so we apply the formula
p_3^20=20!/((20-3)!)
=(20×19×18×17! )/17!=20×19×18
= 6840
Permutations of duplicate items:
The number of permutations of n items, where n1 items are identical, n2 items are identical, n3 items are identical, and so on, is given by:
n!/(n1! n2! n3!?)
Example
In how many distinct ways can the letters of the word MISSISSIPPI be arranged?

The word contains 11 letters (n = 11) where four I are identical (n1 = 4), four S are identical (n2 = 4) and n2 P are identical (n3 = 2). The number of distinct permutations is:

11!/(4! 4! 2!)= (11 .10 .9 .8 .7 .6 .5 .4!)/(4! .4.3 .2.1.2 .1) = 34,650

Example
How many different ways can the letters of the word BANANA be arranged?

There are 60 ways to arrange the letters.
Your turn
How many different ways can the letters of the word ALGEBRA be arranged?
Answer : 2520 ways
combinations
A combination is a way of selecting several things out of a larger group, where (unlike permutations) order does not matter.
A combination of items occurs when
The items are selected from the same group.

No item is used more than once.

The order of items makes no difference.

Remarks
Permutation problems involve situations in which order matters.
Combination problems involve situations in which the order of items makes no difference.
The number of possible combinations if r items are taken from n items is:
C(n,r) = n! /r! (n – r)!

Example
From a class of 20 students we need to select 3 for a committee.
In this case n = 20, and r = 3. The order is not important, so we apply the formula

C_3^20=20!/(3!(20-3)!)=(20?19?18?17!)/(3?2?1?17!)
=1,140

Example
Five cousins at a family reunion decide that three of them will go to pick up a pizza. How many ways can they choose three people to go?
Since the order they choose is not important, you must find the number of combinations of 5 cousins taken three at a time.

C_r^n=n!/r!(n-r)!
=5!/(3!(5-3)!) n=5 ,r=3
=5!/3!2!=10
There are 10 ways to choose three people from the five cousin.


Your turn
Six friends at a party decide that three of them will go to pick up a movie. How many ways can they choose three people to go?
Answer :20 ways

Multiple Events
In more complicated situations, you can use the Fundamental Counting Principle in conjunction with permutations/combinations to determine the number of possibilities

For example, suppose I have a class with 15 boys and 10 girls and I want to send 2 boys and 2 girls to represent our Algebra class at the U.N.

Do C(15,2) for the boys and C(10,2) for the girls, then multiply these answers together!
Example
A soccer club has 8 femal and 7 male members for today s match,the coach want to have 6 femal and 5 meal players on the grass.How many possible configuration are there?
C(8,6) . C(7,5) =8!/(6! 2!)×7!/(5! 2!)
=28×21
=588