Let $A=\{1,3,4,6,9\}$ and $B=\{2,4,5,8,10\}$. Let $R$ be a relation defined on $A imes B$ such tha

Question

Let $A=\{1,3,4,6,9\}$ and $B=\{2,4,5,8,10\}$. Let $R$ be a relation defined on $A 	imes B$ such that $R=\left\{\left(\left(a_{1}, b_{1}ight)ight.ight.$, $\left.\left(a_{2}, b_{2}ight)ight): a_{1} \leq b_{2}$ and $\left.b_{1} \leq a_{2}ight\}$. Then the number of elements in the set $R$ is\

Professor PC Thomas Classes · Answer

(d): We have, $A=\{1,3,4,6,9\} ; B=\{2$, $4,5,8,10\}$\ $R=\left\{\left(\left(a_{1}, b_{1}ight),\left(a_{2}, b_{2}ight)ight): a_{1} \leq b_{2}ight.$ and $\left.b_{1} \leq a_{2}ight\}$\ At $a_{1}=1$, there are 5 choices for $b_{2}$;\ $a_{1}=3$, there are 4 choices for $b_{2}$;\ $a_{1}=4$, there are 4 choices for $b_{2}$;\ $a_{1}=6$, there are 2 choices for $b_{2}$;\ $a_{1}=9$, there are 1 choice for $b_{2}$;\ $	herefore \quad$ Total ways for $\left(a_{1} \leq b_{2}ight)=16$\ Now, at $b_{1}=2$, there are 4 choices for $a_{2}$;\ $b_{1}=4$, there are 3 choices for $a_{2}$;\ $b_{1}=5$, there are 2 choices for $a_{2}$;\ $b_{1}=8$, there are 1 choice for $a_{2}$\ Total ways for $\left(b_{1} \leq a_{2}ight)=10$\ $	herefore$ Required number ways $=16 	imes 10=160$