If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is

The correct option is 0.

To understand why this is correct, let us analyze the conditions required for a mapping (or function) to be both one-one (injective) and onto (surjective) from a finite set A to another finite set B. A mapping that is both one-one and onto is called a bijection.

Let the number of elements in set A be:
$n\left(A\right)=5$
And the number of elements in set B be:
$n\left(B\right)=6$

For a mapping from A to B to be one-one (injective), distinct elements in set A must map to distinct elements in set B. This requires:
$n\left(A\right)\le n\left(B\right)$
Here, $5\le 6$, so one-one mappings are possible.

For a mapping to be onto (surjective), every element in the codomain set B must be mapped to by at least one element in the domain set A. By definition, a function cannot map a single element of A to multiple elements of B. Therefore, to cover all elements in B, there must be at least as many elements in A as there are in B. This requires:
$n\left(A\right)\ge n\left(B\right)$

However, in this case, we have:
$n\left(A\right)<n\left(B\right)$
since $5<6$. This means that any mapping from A to B will leave at least one element in B without a pre-image in A, making an onto mapping impossible to construct.

Since we cannot form any onto mappings from A to B, the number of mappings that are both one-one and onto is:
$0$