The total number of functions, f : $\{1,2,3,4\} \to \{1,2,3,4,5,6\}$ such that $\mathrm{f}(1)+\mathrm{f}(2)=\mathrm{f}(3)$, is equal to :

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Sets Relations and Functions

The total number of functions, f :

{1, 2, 3, 4} \to {1, 2, 3, 4, 5, 6}

such that

f (1) + f (2) = f (3)

, is equal to :

Answer: (B)

Let

A = {1, 2, 3, 4}

and

B = {1, 2, 3, 4, 5, 6}

. The constraint

f (1) + f (2) = f (3)

means

f (3)

can be

2, 3, 4, 5, 6

. For each value of

f (3)

, we count the ordered pairs

(f (1), f (2))

that sum to it, then multiply by 6 (the choices for

f (4)

$f (3) = 2$ : $(1, 1)$ → 1 pair × 6 = 6 cases
$f (3) = 3$ : $(1, 2), (2, 1)$ → 2 pairs × 6 = 12 cases
$f (3) = 4$ : $(1, 3), (3, 1), (2, 2)$ → 3 pairs × 6 = 18 cases
$f (3) = 5$ : $(1, 4), (4, 1), (2, 3), (3, 2)$ → 4 pairs × 6 = 24 cases
$f (3) = 6$ : $(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)$ → 5 pairs × 6 = 30 cases

Total =

6 + 12 + 18 + 24 + 30 = 90

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option