Let A=1,3,4,6,9 and B=2,4,5,8,10.

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 $R={(({a}_{1},{b}_{1}),({a}_{2},{b}_{2})):{a}_{1}\le {b}_{2}$ and ${b}_{1}\le {a}_{2}}$. Then the number of elements in the set R is :

Accepted Answer

Given that the sets are $A={1,3,4,6,9}$ and $B={2,4,5,8,10}$, for the relation $R$ on the set $A	imes B$, we need to find the combinations of pairs that satisfy the conditions ${a}_{1}\le {b}_{2}$ and ${b}_{1}\le {a}_{2}$.

We find the number of combinations by considering the possible values for ${b}_{2}$ for each ${a}_{1}$ and the possible values for ${a}_{2}$ for each ${b}_{1}$ :

For each ${a}_{1}$ in $A={1,3,4,6,9}$, the number of valid ${b}_{2}$ values in $B={2,4,5,8,10}$ are :

- For ${a}_{1}=1$, there are 5 choices for ${b}_{2}$.

- For ${a}_{1}=3$, there are 4 choices for ${b}_{2}$.

- For ${a}_{1}=4$, there are 4 choices for ${b}_{2}$.

- For ${a}_{1}=6$, there are 2 choices for ${b}_{2}$.

- For ${a}_{1}=9$, there is 1 choice for ${b}_{2}$.

This results in a total of $5+4+4+2+1=16$ possible pairs $({a}_{1},{b}_{2})$.

Similarly, for each ${b}_{1}$ in $B$, the number of valid ${a}_{2}$ values in $A$ are :

- For ${b}_{1}=2$, there are 4 choices for ${a}_{2}$.

- For ${b}_{1}=4$, there are 3 choices for ${a}_{2}$.

- For ${b}_{1}=5$, there are 2 choices for ${a}_{2}$.

- For ${b}_{1}=8$, there is 1 choice for ${a}_{2}$.

- For ${b}_{1}=10$, there are no choices for ${a}_{2}$.

This results in a total of $4+3+2+1+0=10$ possible pairs $({b}_{1},{a}_{2})$.

Therefore, the total number of elements in the relation $R$, which satisfies the given conditions, is $16	imes 10=160$.

So, the correct answer is 160.

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