Question

c= {2,3,4,6,8} F={1,2,5,8,9} show that n (cnf) = n(c)+n(f) -n (c) f ​

Answer 1

Step-by-step explanation:

Given Problem:-

c= {2,3,4,6,8} F={1,2,5,8,9} show that n (cnf) = n(c)+n(f) -n (c) f

Correct Question:-

C= {2,3,4,6,8} F={1,2,5,8,9} show that n(CnF) = n(C)+n(F) -n (CUF)

Solution:-

Given that:-

C={2,3,4,6,8}

n(C)=5

F={1,2,5,8,9}

n(F)=5

CUF={2,3,4,6,8}U{1,2,5,8,9}={1,2,3,4,5,6,8,9}

n(CUF)=8

CnF={2,3,4,6,8}n{1,2,5,8,9}={2,8}

n(CnF)=2

LHS:-

n(CnF)=2---------(1)

Now RHS:-

n(C)+n(F) -n (CUF)

=>5+5-8

=>10-8

=>2

n(C)+n(F) -n (CUF)=2-----(2)

From (1)&(2)

n(CnF) = n(C)+n(F) -n (CUF)

Hence, Proved.

Answer:-

n(CnF) = n(C)+n(F) -n (CUF)

Used formula:-

Fundamental Theorem on sets:-

"If A ,B,C are any three sets then

n(AUB)=n(A)+n(B)-n(AnB) or

n(AnB)=n(A)+n(B)-n(AUB)".