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 4: many possibilities: combinations

Many possibilities
combinations

Many choices: Combinations

We want to choose 3 books among the followings:
-- 5 books of Mathematics,
-- 4 books of Physics, and
-- 2 books of Biology.
How many possibilities do we have?

If we choose one book of Mathematics, we will have 4 choices for the Physics book, and for each possibility, we will have 2 other possibilities to choose a book of Biology.

C(3,5) x C(0,4) x (0,2) = 10 x 1 x 1 = 10 possibilities
C(2,5) x C(1,4) x (0,2) = 10 x 4 x 1 = 40 possibilities
C(2,5) x C(0,4) x (1,2) = 10 x 1 x 2 = 20 possibilities
C(1,5) x C(2,4) x (0,2) = 5 x 6 x 1 = 30 possibilities
C(1,5) x C(1,4) x (1,2) = 5 x 4 x 2 = 40 possibilities
C(1,5) x C(0,4) x (2,2) = 5 x 1 x 1 = 5 possibilities

C(0,5) x C(3,4) x (0,2) = 1 x 4 x 1 = 4 possibilities
C(0,5) x C(2,4) x (1,2) = 1 x 6 x 2 = 12 possibilities
C(0,5) x C(1,4) x (2,2) = 1 x 4 x 1 = 4 possibilities

The total is: 165 possibilities

Other opproach:
---------------
The total number of books is 5 + 4 + 2 = 11 books.
The number of combinations of 3 books among 11 is:
C(3,11) = 165 possibilities.

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