ARS COMBINATORIA
| HINDI | symbol | Symbol Name | Meaning / definition | Example |
| | n! | factorial | n! = 1⋅2⋅3⋅…⋅n | 5! = 1⋅2⋅3⋅4⋅5 = 120 |
| | nPk | permutation | _{n}P_{k}=\frac{n!}{(n-k)!} | 5P3 = 5! / (5-3)! = 60 |
| | nCk | combination | _{n}C_{k}=\binom{n}{k}=\frac{n!}{k!(n-k)!} | 5C3 = 5!/[3!(5-3)!]=10 |
Art of combinations and permutations. We can get 3 combinations with 2 objects. lets suppose we have 2 objects called {A , B } , we can get 3 combinations with repetition of AA , BB and one the real combination AB .
When we have 3 objects {A,B,C) and you choose only two objects to make those combinations you will get 6 Combinations with repetition {AA,AB/BA,AC/CA,BB,BC/CB,CC } and without repetition you will get only 3 { AB/BA,AC/CA,BC/CB }
ART OF COMBINATIONS WITH RESPECT TO CONFIGURATIONS
