Q.Find the number of all one-one functions from set $A = \{1, 2, 3\}$ to itself.

Concept: Counting Functions — A one-one (injective) function from a set to itself is a permutation. For a set with $n$ elements, the number of one-one functions is $n!$.

Reasoning:

1. Set $A$ has $3$ elements. A one-one function from $A$ to $A$ must map each element to a distinct image. …

The Concept: Counting Functions as Assignments

Think of a function $f: A \to A$ as a rule that assigns to each element of $A$ some element of $A$ (possibly the same one). For a one-one (injective) function, no two different elements of $A$ can be assigned the same value. Since the domain and codomain are the same finite set, an injective function from a set to itself is automatically onto as well — it's a bijection, also called a permutation of the set.

So the question reduces to: In how many ways can we rearrange the three elements $\{1,2,3\}$?

Step-by-Step

1. Choose the image of 1.

   The element $1$ in the domain can map to any of the 3 elements in the codomain: $1, 2,$ or $3$.

   So there are $3$ choices for $f(1)$.

2. Choose the image of 2, given that $f$ must be one-one.

   Since $f(1)$ is already taken, $2$ cannot map to that same value. That leaves $3 - 1 = 2$ available elements in the codomain.

   So there are $2$ choices for $f(2)$.

3. Choose the image of 3.

   By now, two distinct values have been assigned to $1$ and $2$. Only one element of the codomain remains unused.

   So $3$ must map to that last remaining element — exactly $1$ choice.

4. Multiply the choices. …

Method: Counting One-One Functions with the Multiplication Principle

Use this to count injective functions between finite sets. The core idea: assign images to the domain elements one at a time, and each time an image is used it becomes unavailable, so the number of choices shrinks by one at every step.

Steps

Step 1: Confirm a one-one function can even exist.

A one-one function from an $m$-element set to an $n$-element set needs $m \le n$. When domain and codomain are the same set (size $n$), injective automatically means bijective — a permutation.

Step 2: Count choices element by element.

The first input has all images available; the second must avoid the one already used; the third must avoid two, and so on:

Step 3: Multiply the independent choices. …

Common Mistakes

Mistake 1: Counting all functions ($n^n$) instead of one-one functions.

Why it's wrong: all functions from a $3$-set to itself number $3^3 = 27$, but that allows repeated images; one-one functions forbid repeats. Correct approach: use ${}^{n}P_{n} = n!$, giving $3! = 6$.

Mistake 2: Assuming every function on a set to itself is automatically one-one. …