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
Combinations

Combinations

Combinations

Let's consider another example:
How many ways do we have to choose 3 books among the 5 books: B1, B2, B3, B4, B5 ?

We can choose:
B1, B2, B3
B2, B3, B5
B4, B2, B3
B3, B5, B2
...
and so on

But:
B2, B3, B5 and
B3, B5, B2
form the same combination in which the the order is not important. The repetition of sets is then not allowed when dealing with combinations.

All the combinations we have are:
B1, B2, B3
B1, B2, B4
B1, B2, B5
B1, B3, B4
B1, B3, B5
B1, B4, B5
B2, B3, B4
B2, B3, B5
B2, B4, B5
B3, B4, B5
Their number is 10.

For each combination such as the first one: B1, B2, B3, we have their 3! corresponding permutations:
B1, B2, B3
B1, B3, B2
B2, B1, B3
B2, B3, B1
B3, B1, B2
B3, B2, B1

and so on for the 9 remaining sets.

For each set, we know that the total number of permutions of its three elements is equal to 3!

We know that the total number of arrangements of three elements among the 5 elements is A(3,5) = 5!/(5 -3)!

Then the number of combinations is equal to the number of arrangements divided by the number of permutations. Thus:
C(3,5) = A(3,5)/3! = 5!/3!(5 -3)!

The general case leads to the formula :
The number of combinations of "m" elements among "n" elements is:

C(m,n) = A(m,n)/m! = n!/m!(n - m!
To combine elements = to arrange them without permutations.

Example: We have C(6,49) = 49!/6! (49 - 6)! = 49!/6! 43! = 44 x 45 x 46 x 47 x 48 x 49/6! = 13,983,816 possibilities to win win a lottery 6/49.

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