The probability that a randomly chosen one-one function from the set a, b, c, d to the set 1, 2, 3, 4, 5 satisfies f(a) + 2f(b) - f(c) = f(d) is :

Solution

Maths sir · 4 steps

Fix f(b) first, because it carries the factor 2. Then hunt for c and d with the right sum.

1. Total functions. A one-one function picks 4 different values from 5 for a, b, c, d in order. That gives 5 × 4 × 3 × 2 = 120 functions in all.
2. Rewrite the condition. The condition is f(a) + 2f(b) = f(c) + f(d), with all four values different. Fix f(b), try each value of f(a), and look for two of the remaining numbers that add up to the left side. Each pair gives 2 functions, because f(c) and f(d) can swap.

   f(b)	Working case	{f(c), f(d)}	Functions
   1	f(a) = 5: 5 + 2 = 7	{3, 4}	2
   2	f(a) = 4: 4 + 4 = 8	{3, 5}	2
   3	f(a) = 1: 1 + 6 = 7	{2, 5}	2
   4 or 5	left side is at least 9, too big for two of the remaining numbers	none	0

3. Check the rest. Every other choice of f(a) fails. For example, f(b) = 1, f(a) = 3 needs f(c) + f(d) = 5 from {2, 4, 5}, and no two of these add to 5. So there are 2 + 2 + 2 = 6 favourable functions.
4. So. The probability is 6/120 = 1/20, which is option (d).

The idea to keep. For counting one-one functions under a condition, fix the variable with the biggest coefficient first. Remember that swapping two values in a symmetric sum gives a new function.

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