Cardinality of a Set

Cardinality of a set (n) is defined as the number of elements in a set. Suppose A={2,4,6,7,8}, then n=5.

If A1,A2,A3,A4,.......,Am are the sets with cardinality n_{i} where
i=1,2,3,.......m respectively, then

n(A1∪A2) = n_{1}
(A1) + n_{2}
(A2) - n(A1∩A2)
.........
no. of entries 3

n(A1∪A2∪A3) = n_{1}
(A1) + n_{2}
(A2)
+ n_{3}
(A3) - n(A1∩A2)- n(A2∩A3)- n(A1∩A3)
+ n(A1∩A2∩A3)
.........
no. of entries 7

n(A1∪A2∪A3∪A4)
= n_{1}
(A1) + n_{2}
(A2)
+ n_{3}
(A3)
+ n_{3}
(A4) - n(A1∩A2)- n(A2∩A3)- n(A1∩A3)
- n(A3∩A4)- n(A2∩A4)- n(A1∩A4)+ n(A1∩A2∩A3)
+ n(A2∩A3∩A4)+ n(A1∩A3∩A4)

+ n(A1∩A2∩A4) + n(A1∩A2∩A3∩A4) ......... no. of entries 15

n(UAi) = Σ n(Ai) - Σ n(Ai
∩Aj) + Σ n(Ai
∩Aj ∩Ak) - Σ n(Ai
∩Aj ∩Ak ∩Al)+ ............ +(-1)^{m-1} n( Ai
∩Aj ∩Ak ∩Al.........∩Am)

i= 1 to m i= 1 to m i,j= 1 to m, i<j i,j,k= 1 to m, i<j <k i,j,k,l= 1 to m, i<j <k<l

Observation :

(1) When 2 sets are there, no. of entries 3, when 3 sets,no. of
entries 7, when 4 sets, no.of entries , So when m sets,
Total no. of entries=2^{m} -1. If
one includes the set itself, it becomes 2^{m}

(2) If one studies the distributions, one finds

(a) first is the sum of cardinalities of individual sets, Which is ^{m}C_{1}.

(b) second is sum of cardinalities of various dual combinations of sets,
Which is ^{m}C_{2} with -Ve sign

(c) third is sum of cardinalities of various triple combinations of sets,
Which is ^{m}C_{3}With +Ve sign

_{
(d) so on with alternate changes of sign}

_{
(e) Last is the single combination of all sets }
Which is ^{m}C_{m
}

**Therefore
the total no. of entries is ^{m}C_{1}+ ^{m}C_{2}
+ ^{m}C_{3} + ...... + ^{m}C_{m} = **

1 means one set, 2 means dual combination of sets , 3 means triple combination of sets and so on so forth.

**Table**