Let A = {a, b, c} and B = {1, 2, 3, 4}. Then the number of elements in the set
C = f : A → B | 2 ∈ f(A) and f is not one-one is ______.

Solution

Maths sir · 4 steps

Count the maps that hit 2, then throw away the one-one ones among them.

1. Maps that hit 2. There are 4³ = 64 functions from A to B in all. The ones that never use 2 send each of a, b, c into {1, 3, 4}, which gives 3³ = 27. So 64 - 27 = 37 functions have 2 ∈ f(A).
2. One-one maps hitting 2. There are 4 × 3 × 2 = 24 one-one functions in all. Those that avoid 2 use only {1, 3, 4}, which gives 3 × 2 × 1 = 6. So 24 - 6 = 18 one-one functions have 2 in the range.

   n(C) = 37 - 18 = 19

3. Quick check. You can also count directly. The constant function f ≡ 2 gives 1. If 2 is the repeated value, choose the pair in 3 ways and the third value in 3 ways, giving 9. If 2 is the single value, choose its element in 3 ways and the repeated value in 3 ways, giving 9. The total is 1 + 9 + 9 = 19.
4. So. The set C has 19 elements.

The common slip: 13. Subtracting all 24 one-one functions from 37 gives 13. Only the 18 one-one functions that actually hit 2 belong to that pool.

The idea to keep. When a count has two conditions, handle one with a complement and then remove the unwanted cases from that same smaller pool, never from the full set.

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