MA2102
Probability and Statistics
Lecture-2
Combinations:
Let $r,n$ are non-negative integers, a $r$-combination of n objects is an unordered selection of $r$ of $n$ objects. It can also be called combination of $n$ objects taken $r$ at a time.A $r$-combination of $n$ object simply means a subset of size $r$ from set of $n$ objects.
If , then
-combinations of are
-combinations of are
-combinations of are and there are no 4-combinations as not possible, and.
there is a -combination of S (empty one).
Notation:- number of -combinations of objects. These are called binomial coefficient because of the role it played in binomial theorem(we will see later)
From above example we can see that ,, , and
Theorem: For , , Hence
proof:
Let be set of an objects. Each permutation of can be made by performing following procedure that is broken down into sequence of two tasks,
task-1: choose r objects from S,
task-2: arrange the chosen r objects in some order.
and here task-1 can be done in ways, and task-2 can be done in ways, so this procedure can be done in ways.
Note:
Example: Twenty five points are chosen in the plane out of which 8 points are collinear and no three points come from outside these 8 points are collinear.How many lines do they determine? How many triangle do they determine?
solution: we know that any two points decides unique line, so we are interested in counting point subsets of given twenty five points, there are point subsets ,but point subsets give same line (as 8 of 25 points are collinear ) so the answer is
when it comes to forming a triangle we need point subsets of given twenty give points that do not form a triangle and point subsets of points from triangle only when all come from particular set of points, so will not be correct, but can be fixed as we know exactly of these won't make triangle, hence these points determine