JEE Main 2026 — Mathematics Algebra

Given $A = \{1, 2, 3, 4, 5, 6\}$A={1,2,3,4,5,6}. Since $f: A \rightarrow A$f:A→A is a one-one function, $f(x)$f(x) takes distinct values from $A$A for each $x \in A$x∈A.

We are given the condition $f(2) + f(3) = 5$f(2)+f(3)=5. The possible pairs for $(f(2), f(3))$(f(2),f(3)) from the set $A$A are:

$$(1, 4), (2, 3), (3, 2), (4, 1)$$(1,4),(2,3),(3,2),(4,1)

Notice that for all these pairs, the condition $f(3) \leq 4$f(3)≤4 is automatically satisfied.

We also have the condition $f(1) \geq 3$f(1)≥3. Since $f$f is one-one, $f(1)$f(1) cannot take the values already assigned to $f(2)$f(2) and $f(3)$f(3). Let us find the number of choices for $f(1)$f(1) in each case:

Case 1: $(f(2), f(3)) = (1, 4)$(f(2),f(3))=(1,4)

$f(1) \geq 3$f(1)≥3 and $f(1) \notin \{1, 4\} \Rightarrow f(1) \in \{3, 5, 6\}$f(1)∈/{1,4}⇒f(1)∈{3,5,6}. (3 choices)

Case 2: $(f(2), f(3)) = (2, 3)$(f(2),f(3))=(2,3)

$f(1) \geq 3$f(1)≥3 and $f(1) \notin \{2, 3\} \Rightarrow f(1) \in \{4, 5, 6\}$f(1)∈/{2,3}⇒f(1)∈{4,5,6}. (3 choices)

Case 3: $(f(2), f(3)) = (3, 2)$(f(2),f(3))=(3,2)

$f(1) \geq 3$f(1)≥3 and $f(1) \notin \{3, 2\} \Rightarrow f(1) \in \{4, 5, 6\}$f(1)∈/{3,2}⇒f(1)∈{4,5,6}. (3 choices)

Case 4: $(f(2), f(3)) = (4, 1)$(f(2),f(3))=(4,1)

$f(1) \geq 3$f(1)≥3 and $f(1) \notin \{4, 1\} \Rightarrow f(1) \in \{3, 5, 6\}$f(1)∈/{4,1}⇒f(1)∈{3,5,6}. (3 choices)

In each of the 4 cases, there are exactly 3 choices for $f(1)$f(1).

After assigning values to $f(1), f(2),$f(1),f(2), and $f(3)$f(3), there are exactly 3 elements left in the codomain to be assigned to $f(4), f(5),$f(4),f(5), and $f(6)$f(6). Since $f$f is one-one, these can be assigned in $3! = 6$3!=6 ways.

Total number of such functions = (Number of pairs) $\times$× (Choices for $f(1)$f(1)) $\times$× (Arrangements for remaining elements)

Total = $4 \times 3 \times 6 = 72$4×3×6=72

Answer: $72$72