Many choices: Combinations

We have 5 different books of Mathematics {B1, B2, B3, B4, B5}. How many possibilities do we have to choose 3 books from them? The order is not important.

In other words: Choose 3 books among 5 books!

For B1: 12 possibilities:
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B1 --> B2 --> B3 or B4 or B5 : 3 possibilities
B1 --> B3 --> B4 or B5 : 2 possibilities ( the third possibility with B2
(B1 --> B3 --> B2) is excluded)
B1 --> B4 --> B5 : 1 possibility

B2 --> B3 --> B4 or B5 : 2 possibilities
B2 --> B4 --> B5 : 1 possibility

B3 --> B4 --> B5 : 1 possibility

Finally, we have 3 + 2 + 1 + 2 + 1 + 1 = 10 possiblities.
= 5!/3!(5 - 3)! = (5 different books)!/(3 books among them)![(5 different books)! - (3 books among them)!]! = C(3,5) = combination of 3 books among 5 books.