Tag: cartesian product of sets

Given, $n\left( A \right) =4$ and $n\left( B \right) =5$ 

Product of two sets is the set of ordered pairs formed by mapping every element from the first set to every element of the second set.

So, $ A \times B = $ { $ (1,2), (1,3), (1,4), (2,2), (2,3), (2,4) $ }

And $ B \times A = $ { $ (2,1), (2,2), (3,1), (3,2), (4,1), (4,2) $ }

Clearly, $ A \times B \neq B \times A $

Given $n(A)=3,$  and $n(B) =3$
Hence $n(A\times B) = 3\times 3=9$

If $A$ and $B$ are any two non-empty sets.
then $A\times B$$=\left{(x,y):x\in A  and  y\in B\right}$
As $A = \left{2,3\right}$ and $B = \left{1,2\right}$
$A \times B$$=\left{(2,1), (2,2), (3,1), (3,2)\right}$
Hence, option A.

$\displaystyle A=\left{ 2,4,5 \right} ,B=\left{ 7,8,9 \right} $

$n(A\times B)=n(A)\times n(B)$

As the ordered pairs are equal,
$ 3p + q = p - q $
$  => 2p = -2q $
$ => 2p + 2q = 0 $
$ => p + q = 0 $

$ n(P) $ can be of any value as we are not sure of $ n(Q) $
Hence, $ n(P) $ can take any of the given values.

If $A $ and $B$ are two non empty sets, then the Cartesian product $A \times B$ is set of all ordered pairs $(a,b)$ such that $a\in A$ and $b\in B$.