Question

if A=(1,2,3,4) and B=(3,4,5,6) find n(AuB)*(A intersection B)*(A triangle B)​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{A=\{1,2,3,4\}}$

$\textsf{B=\{3,4,5,6\}}$

$\underline{\textbf{To find:}}$

$\mathsf{n(A{\cup}B){\times}n(A{\cap}B){\times}n(A{\triangle}B)}$

$\underline{\textbf{Solution:}}$

$\textsf{A=\{1,2,3,4\}}$

$\textsf{B=\{3,4,5,6\}}$

$\mathsf{A{\cup}B=\{1,2,3,4,5,6\}\;\implies\;n(A{\cup}B)=6}$

$\mathsf{A{\cap}B=\{3,4\}\;\implies\;n(A{\cap}B)=2}$

$\mathsf{A{\triangle}B=(A{\cup}B)-(A{\cap}B)}$

$\mathsf{A{\triangle}B=\{1,2,3,4,5,6\}-\{3,4\}}$

$\mathsf{A{\triangle}B=\{1,2,5,6\}\;\implies\;nA{\triangle}B)=4}$

$\mathsf{Now,}$

$\mathsf{n(A{\cup}B){\times}n(A{\cap}B){\times}n(A{\triangle}B)}$

$\mathsf{=6{\times}2{\times}4}$

$\mathsf{=48}$

$\implies\boxed{\mathsf{n(A{\cup}B){\times}n(A{\cap}B){\times}n(A{\triangle}B)=48}}$

Answer 2

Solution:

A={1,2,3,4}

B={3,4,5,6}

AUB {1, 2, 3, 4, 5, 6}

→n(AUB) = 6

ΑΠΒ = {3,4} = n(ANB) = 2

AAB (AUB) - (ANB) =

AAB = {1, 2, 3, 4, 5, 6} - {3,4}

ΑΔΒ = {1, 2, 5, 6} → nΑΔΒ) = 4

Now,

n(AUB)xn(ANB)xn(A/B)

= 6×2×4

= 48

=> n(AUB) xn(ANB)xn(AAB) = 48