Let R = {a, b, c, d, e} and S = {1, 2, 3, 4}. Total number of onto functions f : R → S such that f(a

Q: Why can't we just say $f(a)$ has 3 choices (2, 3, 4) and then map the rest freely?

That is the most common trap! If you just assign $f(a)$ to $\{2, 3, 4\}$, you lose control over the 'onto' requirement. The remaining elements might not cover the entire set $S$. You must ensure the function remains surjective, which is why we use the Principle of Inclusion-Exclusion or complementary counting.

Q: Why do we split the $f(a)=1$ case into two sub-cases? Can't we just use a formula?

You could, but splitting is safer and more intuitive. When $f(a)=1$, the remaining elements $\{b, c, d, e\}$ have a different 'onto' requirement depending on whether they need to cover $\{2, 3, 4\}$ or $\{1, 2, 3, 4\}$. Splitting makes the logic crystal clear and prevents errors.

Q: Is there a faster way to do this without the Principle of Inclusion-Exclusion?

For small sets like this, PIE is the most robust method. While you could try to list mappings, you would likely miss a case or double-count. Mastering PIE is a superpower for JEE Advanced; it turns complex counting problems into systematic algebraic steps.

Question

Let $R = \{a, b, c, d, e\}$ and $S = \{1, 2, 3, 4\}$. Total number of onto functions $f : R 	o S$ such that $f(a) 
eq 1$, is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the number of onto functions $f: R 	o S$, where $R = \{a, b, c, d, e\}$ and $S = \{1, 2, 3, 4\}$, subject to the constraint $f(a) 
eq 1$.

The set $R$ has $n=5$ elements, and the set $S$ has $m=4$ elements. An onto function (surjection) requires that every element in $S$ is the image of at least one element in $R$.

### The Total Universe of Onto Functions

To find the total number of onto functions, we employ the **Principle of Inclusion-Exclusion**. The formula for the number of onto functions from a set of size $n$ to a set of size $m$ is:

$$N = \sum_{k=0}^{m} (-1)^k \binom{m}{k} (m-k)^n$$

Substituting $n=5$ and $m=4$:

$$N = 4^5 - \binom{4}{1}3^5 + \binom{4}{2}2^5 - \binom{4}{3}1^5$$

Calculating the individual terms:

$$N = 1024 - 4(243) + 6(32) - 4(1)$$
$$N = 1024 - 972 + 192 - 4 = 240$$

There are exactly **240** total onto functions from $R$ to $S$.

### Applying the Restriction

We must exclude functions where $f(a) = 1$. We use complementary counting: the number of valid functions is the total number of onto functions minus those where $f(a) = 1$.

If $f(a) = 1$, we consider two mutually exclusive cases for the remaining elements $\{b, c, d, e\}$:

**Case 1: $a$ is the only element mapping to $1$.**
The remaining $4$ elements must map onto the remaining $3$ elements $\{2, 3, 4\}$. The number of onto functions from a set of $4$ to a set of $3$ is:

$$3^4 - \binom{3}{1}2^4 + \binom{3}{2}1^4 = 81 - 3(16) + 3(1) = 36$$

**Case 2: $a$ maps to $1$, and at least one other element also maps to $1$.**
In this scenario, the set $\{b, c, d, e\}$ must map onto the entire set $\{1, 2, 3, 4\}$. Since the number of elements equals the number of targets, this is equivalent to the number of permutations of $4$ elements:

$$4! = 24$$

### Final Calculation

The total number of functions where $f(a) = 1$ is the sum of these two cases:

$$36 + 24 = 60$$

Subtracting this from our total universe:

$$240 - 60 = 180$$

The total number of onto functions satisfying $f(a) 
eq 1$ is **180**.

Animated Solution for Mathematics - Permutations and Combinations: Let $R = {a, b, c, d, e}$ and $S = {1, 2, 3, 4}$ . Total number of onto functions $f : R \to S$ such that $f (a) \neq = 1$ , is equal to

Enter Numerical Value:

Visualized Solution

Sets $R$ and $S$

Total Onto Functions Formula

Substituting Values

Calculating Total Functions

Complementary Counting Strategy

Analyzing $f (a) = 1$

Case 1: Only $a$ maps to $1$

Calculating Case 1

Case 2: Others also map to $1$

Calculating Case 2

Total for $f (a) = 1$

Final Answer

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Total Universe of Onto Functions

Applying the Restriction

Final Calculation

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Animated Solution for Mathematics - Permutations and Combinations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R={a,b,c,d,e} and S={1,2,3,4}. Total number of onto functions f:R→S such that f(a)=1, is equal to