Consider two sets $A = \{2, 3, 5, 7, 11, 13\}$ and $B = \{1, 8, 27\}$. Let $f$ be a function from $A$ to $B$ such that for every element $b$ in $B$ , there is at least one element $a$ in $A$ such that $f(a) = b$. Then, the total number of such functions $f$ is

CAT 2024 Slot 1 QA Question 12

Multiple choice (+3 / −1) · Algebra · Functions · Try it, then check the answer and solution below.

CAT 2024 Slot 1QAAlgebra • FunctionsModerate

Consider two sets

A = \{2, 3, 5, 7, 11, 13\}

and

B = \{1, 8, 27\}

. Let

f

be a function from

A

B

such that for every element

b

B , there is at least one element

in

such that

f(a) = b

. Then, the total number of such functions

f$ is

⌨️ Shortcuts:[1-4 / A-D] Select[Enter] Check[B] Star[S] Solution

Official Convenor Answer Key: B

Answer and solution

Answer: B) 540

📌 Core Concept

The problem requires counting the number of onto (surjective) functions from set

A

to set

B

. The formula for the number of onto functions from a set with

m

elements to a set with

n

elements is given by the inclusion-exclusion principle:

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

🔢 Step-by-Step Solution

Identify the sizes of sets:

|A| = 6

and

|B| = 3

Apply the inclusion-exclusion formula:

For

k = 0

(-1)^0 \binom{3}{0} (3 - 0)^6 = 1 \times 1 \times 729 = 729

For

k = 1

(-1)^1 \binom{3}{1} (3 - 1)^6 = -1 \times 3 \times 64 = -192

For

k = 2

(-1)^2 \binom{3}{2} (3 - 2)^6 = 1 \times 3 \times 1 = 3

For

k = 3

(-1)^3 \binom{3}{3} (3 - 3)^6 = -1 \times 1 \times 0 = 0

Sum the results:

729 - 192 + 3 + 0 = 540

⚡ 30-Second Shortcut

Alternatively, using Stirling numbers of the second kind

S(6, 3) = 90 , the number of onto functions is &#36;3! \times 90 = 6 \times 90 = 540

🎯 Final Answer

Correct Answer: Option B