The number of functions f: {1, 2, 3, 4} -> {a, b, c}, which are not onto, is:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – JEE-MAIN 2026

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Home Maths SetsQuestion #2

The number of functions

f : {1, 2, 3, 4} \to {a, b, c}

, which are not onto, is:

Clue 1 of 3: Complementary Counting

It is often easier to calculate the total number of functions and subtract the number of functions that satisfy the opposite condition (onto functions).

🧠 Full Solution Path

● Video Walkthrough

Step 1: Calculate Total Number of Functions✦ Active

Let the domain be $A = {1, 2, 3, 4}$ and the codomain be $B = {a, b, c}$ . The number of elements in the domain is $| A | = m = 4$ , and the number of elements in the codomain is $| B | = n = 3$ . The total number of functions from $A$ to $B$ is given by $n^{m}$ .

N_{t o t a l} = | B |^{| A |} = 3^{4} = 81

Step 2: Calculate Number of Onto Functions○ Expand

The number of onto functions from a set of $m$ elements to a set of $n$ elements can be calculated using the Principle of Inclusion-Exclusion:

N_{o n t o} = \sum_{k = 0}^{n} (- 1)^{k} (\binom{n}{k}) (n - k)^{m}

For $m = 4$ and $n = 3$ :

N_{o n t o} = (\binom{3}{0}) (3 - 0)^{4} - (\binom{3}{1}) (3 - 1)^{4} + (\binom{3}{2}) (3 - 2)^{4} - (\binom{3}{3}) (3 - 3)^{4}

Substitute the values:

N_{o n t o} = (1) (3^{4}) - (3) (2^{4}) + (3) (1^{4}) - (1) (0^{4})

N_{o n t o} = (1) (81) - (3) (16) + (3) (1) - (1) (0)

N_{o n t o} = 81 - 48 + 3 - 0 = 36

💡 Teacher's Secret Hint

Remember that $(\binom{n}{k})$ is the binomial coefficient and $0^{0}$ is typically taken as 1 in combinatorial contexts, but $0^{m}$ for $m > 0$ is $0$ .

Step 3: Calculate Number of Non-Onto Functions○ Expand

The number of functions that are not onto is the total number of functions minus the number of onto functions.

N_{n o t_o n t o} = N_{t o t a l} - N_{o n t o}

N_{n o t_o n t o} = 81 - 36 = 45

💡 Teacher's Secret Hint

This method of complementary counting simplifies the problem significantly.

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Maths Question 4 – JEE-MAIN 2026

Clue 1 of 3: Complementary Counting

🧠 Full Solution Path