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

The problem asks for the number of bijective functions from a set \(A\) to itself when \(A\) contains 97 elements. Let's break down the solution step-by-step.

Understanding The Concept of Bijective Function: 

A bijective function is both injective (one-to-one) and surjective (onto). This means every element in the domain is paired with a unique element in the codomain, and every element in the codomain is covered.

Calculating the Number of Bijective Functions:

When set \(A\) has \(n\) elements, the number of bijections (bijective functions) from \(A\) to \(A\) is equal to the number of permutations of \(n\) elements. This is because each element in the domain has to be paired with a unique element in the codomain.

1. The number of permutations of a set with \(n\) elements is \(n!\) (n factorial). This gives us the formula to find the number of bijections.
2. Since the set \(A\) has 97 elements, the number of bijective functions is \(97!\).

Thus, the answer is \(97!\).

Conclusion:

The correct answer is 97!, which represents the total number of bijective functions from set \(A\) to itself.

Option	Explanation
97	Only represents a count, not factorial. Incorrect.
\((97)^2\)	Represents squared value, not permutations. Incorrect.
97!	Correct as it stands for the permutations of 97 elements.
\(2^{97}\)	Exponential form unrelated to permutations. Incorrect.

Tip: For a set containing \(n\) elements, the number of bijective functions from the set to itself is always \(n!\).