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Question: Statement I: Let C and D be two finite sets having 3 and 4 elements respectively. Then the number of...

Statement I: Let C and D be two finite sets having 3 and 4 elements respectively. Then the number of functions from C to D is 81. Statement II: Every function from set C to set D is a subset of C×D.

A

Both Statement I and Statement II are correct.

B

Statement I is correct but Statement II is incorrect.

C

Statement II is correct but Statement I is incorrect.

D

Both Statement I and Statement II are incorrect

Answer

c. Statement II is correct but Statement I is incorrect.

Explanation

Solution

Statement I Analysis:

Let CC and DD be two finite sets with C=m|C| = m and D=n|D| = n. The number of functions from CC to DD is given by nmn^m.

In this case, C=3|C| = 3 and D=4|D| = 4. So, m=3m=3 and n=4n=4.

The number of functions from CC to DD is 43=4×4×4=644^3 = 4 \times 4 \times 4 = 64.

The statement says the number of functions is 81. Since 648164 \neq 81, Statement I is incorrect.

Statement II Analysis:

A function f:CDf: C \to D is a relation from CC to DD such that for every element xCx \in C, there exists a unique element yDy \in D such that (x,y)(x, y) belongs to the relation.

A relation from CC to DD is defined as a subset of the Cartesian product C×DC \times D. The Cartesian product C×DC \times D is the set of all ordered pairs (c,d)(c, d) where cCc \in C and dDd \in D.

The function ff can be represented by its graph, which is the set of ordered pairs {(c,f(c))cC}\{(c, f(c)) \mid c \in C\}.

For any cCc \in C, f(c)f(c) is the unique element in D assigned to cc. Thus, f(c)Df(c) \in D.

Therefore, each ordered pair (c,f(c))(c, f(c)) is an element of C×DC \times D.

The set {(c,f(c))cC}\{(c, f(c)) \mid c \in C\}, which represents the function ff, is a subset of C×DC \times D.

Thus, every function from set CC to set DD is a subset of C×DC \times D. Statement II is correct.

Conclusion:

Statement I is incorrect. Statement II is correct.

This matches option (c).