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Probability and Statistics



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 S={a,b,c}S=\{a,b,c\}, then
11-combinations of SS are {a},{b},{c}\{a\},\{b\},\{c\}
22-combinations of SS are {a,b},{a,c},{b,c}\{a,b\},\{a,c\},\{b,c\}
33-combinations of SS are {a,b,c}\{a,b,c\} and there are no 4-combinations as not possible, and.
there is a 00-combination of S (empty one).
Notation:- (nr):=\binom{n}{r} :=number of rr-combinations of nn 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 (30)=1\binom{3}{0}=1,(31)=3\binom{3}{1}=3, (32)=3\binom{3}{2}=3,(33)=1\binom{3}{3}=1 and (34)=0\binom{3}{4}=0

Theorem: For 0rn0\le r\le n, nPr=r!(nr)^nP_r=r!\binom{n}{r}, Hence (nr)=n!r!(nr)!\binom{n}{r}=\frac{n!}{r!(n-r)!}
Let SS be set of an nn objects. Each rr- permutation of SS 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 (nr)\binom{n}{r} ways, and task-2 can be done in r!r! ways, so this procedure can be done in (nr)r!\binom{n}{r}r! ways.
 nPr=(nr)r!(nr)=n!r!(nr)!\therefore~ ^nP_r=\binom{n}{r}r! \Rightarrow \binom{n}{r}=\frac{n!}{r!(n-r)!}
Note: (nr)=(nnr)\binom{n}{r}=\binom{n}{n-r}

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 22 point subsets of given twenty five points, there are (252)\binom{25}{2} 22 point subsets ,but (82)\binom{8}{2} 22 point subsets give same line (as 8 of 25 points are collinear ) so the answer is (252)(82)+1\binom{25}{2}-\binom{8}{2} +1
when it comes to forming a triangle we need 33 point subsets of given twenty give points that do not form a triangle and 33 point subsets of points from triangle only when all come from 88 particular set of points, so (253)\binom{25}{3} will not be correct, but can be fixed as we know exactly (253)\binom{25}{3} of these won't make triangle, hence these 2525 points determine (253)(83)\binom{25}{3}-\binom{8}{3}