JEE Advanced 2026 — Paper 2 — MATH-III solution

JEE Advanced2026Paper 2MATH-III

Let $N$ denote the set of all positive integers. Consider the sets

$A = {1, 2, 3, 4, 5}$ and $B = {1, 2, 3, 4, 5, 6, 7}$ .

Let $S$ be the set of all functions $f : A \to B$ such that $f (2) \neq = 2$ and $f (4) \neq = 4$ . Consider the set

$T = {f \in S : there exists a function g : B \to N such that g (f (x)) = 2^{x} for all x \in A}$ .

Then the number of elements in the set $T$ is ___________.

Your answer

Solution

The condition $g (f (x)) = 2^{x}$ for all $x \in A$ forces $f$ to be one-one on $A$ , because distinct $x$ values give distinct outputs $2^{x}$ , which would require distinct $f (x)$ .

Conversely, any injective $f : A \to B$ admits such a $g$ (define $g$ on the image of $f$ via $g (f (x)) = 2^{x}$ and arbitrarily elsewhere into $N$ ). So $T$ is the set of injective $f : A \to B$ with $f (2) \neq = 2$ and $f (4) \neq = 4$ .

Total injective functions from $A$ to $B$ : $(5 7) \cdot 5! = 21 \cdot 120 = 2520$ .

Use inclusion-exclusion on the forbidden sets $E_{2} = {f injective : f (2) = 2}$ and $E_{4} = {f injective : f (4) = 4}$ .

$∣ E_{2} ∣ =$ (fix $f (2) = 2$ , then choose an injection from the remaining 4 elements of $A$ to the remaining 6 elements of $B$ ) $= (4 6) \cdot 4! = 15 \cdot 24 = 360$ . By symmetry $∣ E_{4} ∣ = 360$ .

$∣ E_{2} \cap E_{4} ∣ =$ (fix $f (2) = 2, f (4) = 4$ , inject the remaining 3 elements of $A$ into the remaining 5 of $B$ ) $= (3 5) \cdot 3! = 10 \cdot 6 = 60$ .

$∣ T ∣ = 2520 - 360 - 360 + 60 = 1860$ .

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