The number of bijective functions $f :\{1,3,5$, $7, \ldots \ldots 99\} \rightarrow\{2,4,6,8, \ldots \ldots , 100\}$, such that $f (3) \geq f (9) \geq f (15) \geq f (21) \geq \ldots f (99), $ is _____
- ${ }^{50} P _{17}$
- ${ }^{50} P _{33}$
- $33 ! \times 17 !$
- $\frac{50 \text { ! }}{2}$
The Correct Option is B
Solution and Explanation
numbers
for condition one we have way rest elements
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Concepts Used:
Relations and functions
A relation R from a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.
Representation of Relation and Function
Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, represent this function in different forms.
- Set-builder form - {(x, y): f(x) = y2, x ∈ A, y ∈ B}
- Roster form - {(1, 1), (2, 4), (3, 9)}
- Arrow Representation
