Question

The number of bijective functions from set $A$ to itself when $A$ contains $97$ elements, is:

• A. 97
• B. \((97)^2\)
• C. 97!
• D. \(2^{97}\)

Hint:

If a set has \(n\) elements, the total number of functions is \(n^n\), but only \(n!\) of those are bijections. As \(n\) grows, bijections become a very small fraction of the total possible functions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A bijective function from a finite set to itself is a permutation of the elements of that set. Every element in the domain must map to a unique element in the codomain so that no two elements share a target and no target is left unmapped.

Step 2: Key Formula or Approach:
The number of bijections from a set with \(n\) elements to another set with \(n\) elements is given by \(n!\) (n-factorial).

Step 3: Detailed Explanation:
1. Let set \(A = \{a_1, a_2, \dots, a_{97}\}\). 2. To form a bijection, the first element \(a_1\) can be mapped to any of the 97 elements in the codomain (97 choices). 3. The second element \(a_2\) must be mapped to a unique element, so it has 96 choices remaining. 4. This pattern continues until the last element has only 1 choice left. 5. Total functions = \(97 \times 96 \times 95 \times \dots \times 1 = 97!\).

Step 4: Final Answer
The number of bijective functions is 97!.