The number of relations R from an m -element set A to an n-element set B satisfying the condition (a b1)∈ R , (a, b2)∈ R ⇒ b1=b2 for a∈ A ,b1,b2 ∈ B is

Question

The number of relations R from an m -element set A to an n-element set B satisfying the condition (a b1)∈ R , (a, b2)∈ R ⇒ b1=b2 for a∈ A ,b1,b2 ∈ B is (A) nm (B) 2m+n-2m-2n (C) mn (D) (n+1)m

Tardigrade Admin · Accepted Answer

Set A have m-elements, Set B have n-elements (a b1)∈ R , (a, b2)∈ R ⇒ (b1=b2) By condition relation is a function ∴ Total number of function (Relation) =nm

Q. The number of relations $R$ from an m -element set $A$ to an $n$ -element set B satisfying the condition $(a b_{1}) \in R, (a, b_{2}) \in R \Rightarrow b_{1} = b_{2}$ for $a \in A, b_{1}, b_{2} \in B$ is

$n^{m}$

$2^{m + n} - 2^{m} - 2^{n}$

$mn$

$(n + 1)^{m}$

Set $A$ have $m$ -elements,
Set $B$ have $n$ -elements
$(a b_{1}) \in R, (a, b_{2}) \in R$
$\Rightarrow (b_{1} = b_{2})$
By condition relation is a function
$∴$ Total number of function (Relation) $= n^{m}$

Q. The number of relations R from an m -element set A to an n-element set B satisfying the condition (ab1​)∈R,(a,b2​)∈R⇒b1​=b2​ for a∈A,b1​,b2​∈B is

nm

2m+n−2m−2n

mn

(n+1)m