chalte chalte combinatorics 3

The following are the only possibilities:
i) f(0)=f(1)=f(2)
ii) f(0)<f(1)=f(2)
iii)f(0)=f(1)<f(2)
iv)f(0)<f(1)<f(2)

For iv) we just need to select 3 numbers from B and arrange them in ascending order. This could be done in ⁸C₃ ways.
For ii) and iii), we similarly have ⁸C₂ ways.
For i) there are ⁸C₁ ways.

Hence, the total number of required functions = ⁸C₃ + 2 ⁸C₂ + ⁸C₁ = 120 = ¹⁰C₃

ingenious.....thanks sir!!!