Counting Composite Bijections | Ashish Khare Sir

Counting Composite Bijections | Ashish Khare Sir

In mathematics functions may have discreet domain and codomain, which gives a reason to count the number of possible functions in various cases. These cases are very much used in problem making for IIT JEE Main and Advanced
What is Bijective Function?
A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A.
What are composite functions?

Composite functions are when the output of one function is used as the input of another.
The composition of functions f(x) and g(x) where g(x) is acting first is represented by f(g(x)) or (f ∘ g)(x). It combines two or more functions to result in another function. In the composition of functions, the output of one function that is inside the parenthesis becomes the input of the outside function. i.e.,
• In f(g(x)), g(x) is the input of f(x).
• In g(f(x)), f(x) is the input of g(x).

How do you count bijective functions?
Number of Bijective functions

If there is bijection between two sets A and B, then both sets will have the same number of elements. If n(A) = n(B) = m, then number of bijective functions = m!.

All aspects of Mathematics as in Calculus, Coordinate Geometry, Trigonometry, Algebra, Statistics, Set Theory will be dealt with. These will be focused on the IIT JEE Main and Advanced syllabus. Also, topics like Euclidean Geometry, Combinatorics, Number Theory, Inequalities etc. would be dealt with keeping in mind the International Olympiads. These FUNDAE will be relevant for IOQM, INMO, IMO, SUMaC, PROMYS, IGO, NMTC, ROSS, Tournaments of the Town.

Question TBQ-022(M) - The number of bijective functions f:{1,2,3} →:{1,2,3} such that f(f(f(x))) = x is 4
(A) True (B) False

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[ckjain_maths]

If 1 maps to 1, 2 has to map with 1. Hence, mapping in pairs shall give the desired composite function.
There are six cases.
Case 1--- when no element maps to itself.
Since number of elements are odd. This case will not give desired composition.
Case 2--- when only one element maps to itself. Rest of four element shall make two pairs in which elements of each pair maps each other.
Identical mapping element can be chosen in 5 ways. Two pairs out of four elements can be chosen in 4C2 ways i.e. 6 ways. There are 6*5= 30 ways.
Case 3 --- when only two elements maps to itself. Emphasis is on " only two". Rest three can not the required composite function.
Case 4-- when only three elements maps to itself. 3 elements out of 5 can be chosen in 5C2 = 10 ways.
Case 5 --- when only four elements maps to itself. Required composite function is not possible.
Case 6--- when all five elements maps to itself. One case is possible.
Total number of bijections for desired composite function is 30+10+1= 41.

[chewie]

26 I think
The function will be equal to its inverse. Therefore, if (x, y) is part of f(x) then so is (y, x). Thus the numbers will either exist in pairs or they will map to themselves. We can set up three cases:
0 pairs: all numbers map to themselves, 1 possibility.
1 pair: any two numbers will pair up, i.e. 5C2=10
2 pairs: any one of 5 numbers will map to itself. That is one number pairs to itself and the other 4 can arrange into 3 possible pairs. (4!/2!*2!*2!). Thus, 3*5= 15.
1+10+15=26

[AnshPl]

Sir 26 for the question you gave? Thank you sir, session was amazing
Edit: Sir i did it by recurrence