Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is (a) 14 (b) 16

Q: Why do we subtract the 'into' functions? Can't we just count the 'onto' ones directly?

That is a fantastic question! You absolutely can count them directly, but it requires careful case-by-case analysis. For example, you could have $3$ elements mapping to $1$ and $1$ element mapping to $2$, or $2$ elements mapping to $1$ and $2$ elements mapping to $2$. While this works for small sets, the 'Total - Into' strategy is a powerful, universal heuristic in combinatorics that prevents you from missing cases or double-counting. It turns a complex problem into a simple subtraction!

Q: What if the codomain had $3$ elements instead of $2$? Does the formula $2^n - 2$ still work?

No, the formula $2^n - 2$ is specific to when the codomain has exactly $2$ elements. If the codomain has $m$ elements, the number of onto functions is calculated using the Principle of Inclusion-Exclusion. For $m=3$, you would subtract cases where the range is a subset of size $1$ or $2$. It becomes more involved, but the core logic of 'Total - Bad' remains the same.

Q: Is it possible for an 'into' function to have a range with $2$ elements?

No, by definition, an 'into' function means the range is a *proper* subset of the codomain. Since our codomain $F$ has only $2$ elements, the only proper subsets are sets with $1$ element. Therefore, any 'into' function must have a range with exactly $1$ element. If the range had $2$ elements, it would be the entire codomain, making it an 'onto' function.

Question

Let $E = \{1, 2, 3, 4\}$ and $F = \{1, 2\}$. Then the number of onto functions from $E$ to $F$ is

Options:

(A) $14$

(B) $16$

(C) $12$

(D) $8$

Sigma Percentile · Accepted Answer

### The Elegance of Mapping: Unlocking Onto Functions

Welcome, fellow traveler on the path of JEE mastery! Today, we are going to explore a fundamental concept in set theory and functions: the onto function, or as mathematicians formally call it, the surjective function.

It might seem like a simple counting problem, but beneath the surface lies the beautiful logic of combinatorics. Let us dive in.

### Defining the Battlefield

We are given two sets: the domain $E = \{1, 2, 3, 4\}$ and the codomain $F = \{1, 2\}$. Our task is to find the number of onto functions from $E$ to $F$.

Think of this as a scenario where you have $4$ distinct items (the elements of $E$) that you need to distribute into $2$ distinct bins (the elements of $F$).

### The Surjective Condition

What does it mean for a function $f: E ightarrow F$ to be onto? It means that every element in the codomain $F$ must be 'hit' by at least one element from the domain $E$.

In our bin analogy, it means no bin can be left empty. If even one bin is empty, the function fails to be onto. This is the 'no-one-left-behind' policy of functions.

### The Power of Complementary Counting

Directly counting the onto functions can be tedious. We would have to consider all the different ways to partition $4$ elements into $2$ non-empty sets.

Instead, we use a brilliant strategy: Complementary Counting. We calculate the total number of possible functions and subtract the 'bad' ones—the ones that are not onto, which we call 'into' functions.

$$
	ext{Onto Functions} = 	ext{Total Functions} - 	ext{Into Functions}
$$

### Calculating the Total

First, let's find the total number of functions. For each element in $E$, there are $2$ choices in $F$.

Since there are $4$ elements in $E$, the total number of functions is $2 	imes 2 	imes 2 	imes 2$, or $2^4$.

$$
2^4 = 16
$$

So, there are $16$ possible ways to map $E$ to $F$.

### The 'Into' Trap

Now, what are the 'into' functions? These are the functions where the range is a proper subset of $F$.

Since $F = \{1, 2\}$, the only proper subsets are $\{1\}$ and $\{2\}$. This means an 'into' function occurs if all elements of $E$ map to just $1$, or all elements of $E$ map to just $2$.

Case $1$: All elements map to $1$. There is only $1$ such function.

Case $2$: All elements map to $2$. There is only $1$ such function.

Thus, there are exactly $2$ 'into' functions.

### The Final Victory

Now, we simply apply our subtraction strategy:

$$
	ext{Onto Functions} = 16 - 2 = 14
$$

And there we have it! The number of onto functions is **$14$**.

### A Note for Your Toolkit

This problem highlights a general rule. For a domain of size $n$ and a codomain of size $2$, the number of onto functions is always $2^n - 2$.

Keep this in your mental toolkit; it is a classic JEE Advanced concept that rewards those who understand the underlying logic rather than just memorizing formulas. Keep practicing, keep visualizing, and most importantly, keep falling in love with the process!

Animated Solution for Mathematics - Functions: Let $E = {1, 2, 3, 4}$ and $F = {1, 2}$ . Then the number of onto functions from $E$ to $F$ is

Select Answer:

Visualized Solution

Visualizing Sets $E$ and $F$

What is an Onto Function?

The Subtraction Strategy

Calculating Total Functions

Total Functions $= 16$

Identifying the "Into" Functions

The Two Specific Into Cases

Subtracting to Find Onto Functions

General Formula for Codomain of Size 2

The Sigma Insight: Classification of Functions

The Elegance of Mapping

Defining the Battlefield

The Surjective Condition

The Power of Complementary Counting

Calculating the Total

The 'Into' Trap

The Final Victory

A Note for Your Toolkit

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let E={1,2,3,4} and F={1,2}. Then the number of onto functions from E to F is