Let A = {1,2,3}, B= {1,2,3,4} and C= {1,2,3,4,5}. For the sets S and T described

Question

Let A = {1,2,3}, B= {1,2,3,4} and C= {1,2,3,4,5}. For the sets S and T described below, explain whether |S| < |T|, |S| > |T| or |S| = |T|.

(a) Let B be the universal set and let S be the set of all subsets X of B for which |X| (not equal) |X bar|. Let T be the set of 2-element subsets of C.

(b) Let S be the set of all partitions of the set A and let T be the set of 4-element subsets of C.

(c) Let S = {(b,a) : b element B, a element A, a+b is odd} and let T be the set of all nonempty proper subsets of A.

Explanation / Answer

Its a easy question.

for (a) |S|=|T|,

B={1,2,3,4} is the universal set.

cardinality of X can be 0,1,3,4. now we know number of subsets of a set with elements is 2^n .

in this case we need to discard the subsets with 2 elements so cardinality of S =2^4-4C2=16-6=10,

and cardinality of T is just 5C2=10.

so.

(b)

S is the set of all partition of A .now partition of A can be 3,1+2,1+1+1.

so number of partions is 1+3C1+1=5,

so cardinality is 5.

T is the 4 elements subsets of C which is simply 5C4=5,

so |S|=|T|,

for(C).

we first calculate the cardinality of S,

if b is odd a has to be even.so number of pairs 2.1=2,

if b is even a has to be odd.so number of pairs 2.2=4,

so cardinality is 4+2=6,

obviously cardinality of T is 2^3-2 =6[phi and A are excluded],

so |S|=|T| also.

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