Contents
 • one choice • many choices: permutations • many choices: combinations • many choices: combinations • independent events • dependent events • permutations without repettitions • permutations with identical elements • combinations • coditional probabilities • and or events • two outcomes • total probability theorem • Bayes rule • union sets and probability • fundamental counting principle

 Other exercices

 Combinatorics - Probability
Example 2: many possibilities: permutations
Many possibilities
Permutations

Many choices: Permutations

We have 5 different books of Mathematics. How many possibilities can we have to choose 3 books from them?. The order is important.

At first, we have 5 possibilities to choose a book. If we choose one book, il will remain 4 possibilities to choose another second book from the 4 remaininig books. Finally, it remains 3 possibilities to choose the last book from the 3 remaininig books. Hence, the total possibilities that we have is 5 x 4 x 3 = 60 possibilities.

Other approach:
For B1: 12 possibilities:
-------------------------
B1 --> B2 --> B3 or B4 or B5 :3 possibilities
B1 --> B3 --> B2 or B4 or B5 :3 possibilities
B1 --> B4 --> B2 or B3 or B5 :3 possibilities
B1 --> B5 --> B2 or B3 or B4 :3 possibilities

Similarly for B2, B3, B4, and B5
Finally, we have 5 x 12 = 60 possiblities.
= 3 x 4 x 5 = 5!/(5 - 3)! = (5 different books)!/[(5 different books)! - (3 books among them)!]! = P(3,5) = Permutation of 3 books among 5 books.

Main formula

Let's consider a set of "n" different elements: {B1, B2, B3, ..., Bn}. The arrangement of "p" elements chosen among the "n" elements is the ordered disposition of "p" elements. For example, {B1, B3, B4, B9} is an arrangement of 4 elements among the "n" elements. {B1, B9, B4, B3} is another arrangement. {B1, B3, B4, B9} and {B1, B9, B4, B3} are two different arrangements because the order of the elements is different in the two sets (two arrangements).

At first, we have "n" possibilities to choose an element from the set {B1, B2, B3, ..., Bn}. Then, once the first is choosen, it remains "n - 1" possibilities to choose a second. Then "n - 2" to choose a third element. ... Then "n - (p - 1)" to choose the "pth" element. Finally, the total number of possibilities to arrange "p" elements among "n" elements is :
A(p,n) = n x (n - 1) x (n - 2) x ... x [n - (p - 1)]     (1)

We can write A(p,n) = A(p,n) x R /R     (3)
Where: R = [n - (p )] x [n - (p +1)] x [n - (p + 2)] x ... x 2 x 1 = (n - p)!    (3)

The formula (1) becomes: A(p,n) x R = n!
Hence:

A(p,n) = n!/(n - p)!

The number of permutations of the "n" elements of the set {B1, B2, B3, ..., Bn} we can make is the number of arragements of "n" elements among the "n" elements of the seté. That is: P(n,n) = A(n,n) = n!/(n - n)! = n!

P(n,n) = n!

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