JEE Main 2022 · 25 July · morning · Single correct · +4 / −1

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

  1. (a) 60
  2. (b) 90
  3. (c) 108
  4. (d) 126
Answer and solution

Answer (b) 90

Solution

Maths sir · 3 steps

Fix f(3) first, count the pairs that add up to it, then let f(4) go free.

  1. Split the work. Only f(1), f(2) and f(3) are tied together by the condition. f(4) has no restriction, so it can take any of the 6 values.
  2. Count the pairs. Choose the value of f(3), then count the ordered pairs (f(1), f(2)) that add up to it. Both values are at least 1, so f(3) must be at least 2.
    f(3)Ordered pairs (f(1), f(2))Count
    1none0
    2(1,1)1
    3(1,2), (2,1)2
    4(1,3), (2,2), (3,1)3
    5(1,4), (2,3), (3,2), (4,1)4
    6(1,5), (2,4), (3,3), (4,2), (5,1)5

    (0+1+2+3+4+5) × 6 = 15 × 6 = 90

  3. So. There are 15 valid choices for f(1), f(2), f(3) and 6 choices for f(4), giving 90 functions. The answer is (b).

The idea to keep. When a condition ties only some inputs together, count those cases on their own, then multiply by the choices for the free inputs. Always check that every value you use is in the codomain.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 160 times in JEE Main since 2020. This question is from 2022.