Surjective Functions

An onto (surjective) function covers every codomain element at least once, and counting them uses inclusion-exclusion over codomain elements that might be missed, a hallmark JEE Advanced technique. The total functions from 4 to 3 elements equal 3^4 = 81. Subtract functions missing at least one codomain value: C(3,1)·2^4 = 3·16 = 48. Add back those missing at least two values, over-subtracted: C(3,2)·1^4 = 3·1 = 3. The alternating sum gives 81 - 48 + 3 = 36 onto functions. Option 81 counts all functions, including non-surjective ones. Option 18 halves incorrectly without justification. Option 24 = 4! confuses this with a bijection count, impossible since the sets differ in size. Hence there are 36 surjections. Plausibility check: the surjection count must be smaller than the total 81 and consistent with the Stirling number S(4,3)=6 times 3! = 36, and indeed 6·6 = 36, confirming the inclusion-exclusion result. The inclusion-exclusion count of surjections generalizes to any codomain size, and the resulting alternating sum equals n! multiplied by the Stirling number of the second kind that counts set partitions into the required number of non-empty blocks. Recognizing this bridge between surjective functions and partitions of a finite set is a frequently rewarded structural insight in advanced counting.