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

We need to find the probability that a randomly chosen one-one function $$f: \{a, b, c, d\} \to \{1, 2, 3, 4, 5\}$$ satisfies $$f(a) + 2f(b) - f(c) = f(d)$$.

Count total one-one functions:

$$\text{Total} = 5 \times 4 \times 3 \times 2 = 120$$

Find all injections satisfying the equation:

We need four distinct values from $$\{1, 2, 3, 4, 5\}$$ assigned to $$a, b, c, d$$ such that $$f(a) + 2f(b) - f(c) = f(d)$$.

Equivalently: $$f(a) + 2f(b) = f(c) + f(d)$$, where $$f(a), f(b), f(c), f(d)$$ are all distinct.

Let us denote the values as $$p = f(a), q = f(b), r = f(c), s = f(d)$$, all distinct, with $$p + 2q - r = s$$ and each in $$\{1, 2, 3, 4, 5\}$$.

We systematically check all possible values of $$q$$ (since it has the largest coefficient):

Case $$q = 1$$: $$s = p + 2 - r$$. Try all distinct $$(p, r)$$ from remaining values $$\{2,3,4,5\}$$:

• $$(p, r) = (3, 4)$$: $$s = 3 + 2 - 4 = 1 = q$$ $$\times$$
• $$(p, r) = (4, 3)$$: $$s = 4 + 2 - 3 = 3 = r$$ $$\times$$
• $$(p, r) = (3, 5)$$: $$s = 0 \notin \{1,...,5\}$$ $$\times$$
• $$(p, r) = (5, 3)$$: $$s = 4$$, all distinct: $$\{5, 1, 3, 4\}$$ $$\checkmark$$
• $$(p, r) = (4, 5)$$: $$s = 1 = q$$ $$\times$$
• $$(p, r) = (5, 4)$$: $$s = 3$$, all distinct: $$\{5, 1, 4, 3\}$$ $$\checkmark$$
• $$(p, r) = (2, 3)$$: $$s = 1 = q$$ $$\times$$
• $$(p, r) = (3, 2)$$: $$s = 3 = p$$ $$\times$$
• $$(p, r) = (2, 4)$$: $$s = 0$$ $$\times$$
• $$(p, r) = (4, 2)$$: $$s = 4 = p$$ $$\times$$
• $$(p, r) = (2, 5)$$: $$s = -1$$ $$\times$$
• $$(p, r) = (5, 2)$$: $$s = 5 = p$$ $$\times$$

2 valid assignments for $$q = 1$$.

Case $$q = 2$$: $$s = p + 4 - r$$. Try $$(p, r)$$ from $$\{1,3,4,5\}$$:

• $$(1, 4)$$: $$s = 1$$, $$= p$$ $$\times$$
• $$(4, 1)$$: $$s = 7$$ $$\times$$
• $$(1, 5)$$: $$s = 0$$ $$\times$$
• $$(5, 1)$$: $$s = 8$$ $$\times$$
• $$(3, 4)$$: $$s = 3 = p$$ $$\times$$
• $$(4, 3)$$: $$s = 5$$, all distinct: $$\{4, 2, 3, 5\}$$ $$\checkmark$$
• $$(3, 5)$$: $$s = 2 = q$$ $$\times$$
• $$(5, 3)$$: $$s = 6$$ $$\times$$
• $$(1, 3)$$: $$s = 2 = q$$ $$\times$$
• $$(3, 1)$$: $$s = 6$$ $$\times$$
• $$(4, 5)$$: $$s = 3$$, all distinct: $$\{4, 2, 5, 3\}$$ $$\checkmark$$
• $$(5, 4)$$: $$s = 5 = p$$ $$\times$$

2 valid assignments for $$q = 2$$.

Case $$q = 3$$: $$s = p + 6 - r$$. Try $$(p, r)$$ from $$\{1,2,4,5\}$$:

• $$(1, 2)$$: $$s = 5$$, all distinct: $$\{1, 3, 2, 5\}$$ $$\checkmark$$
• $$(2, 1)$$: $$s = 7$$ $$\times$$
• $$(1, 4)$$: $$s = 3 = q$$ $$\times$$
• $$(4, 1)$$: $$s = 9$$ $$\times$$
• $$(1, 5)$$: $$s = 2$$, all distinct: $$\{1, 3, 5, 2\}$$ $$\checkmark$$
• $$(5, 1)$$: $$s = 10$$ $$\times$$
• $$(2, 4)$$: $$s = 4 = r$$ $$\times$$
• $$(4, 2)$$: $$s = 8$$ $$\times$$
• $$(2, 5)$$: $$s = 3 = q$$ $$\times$$
• $$(5, 2)$$: $$s = 9$$ $$\times$$
• $$(4, 5)$$: $$s = 5 = r$$ $$\times$$
• $$(5, 4)$$: $$s = 7$$ $$\times$$

2 valid assignments for $$q = 3$$.

Case $$q = 4$$: $$s = p + 8 - r$$. Since the minimum of $$p + 8 - r$$ with $$p \geq 1$$ and $$r \leq 5$$ is $$1 + 8 - 5 = 4$$, most values will be too large or equal to $$q = 4$$. Checking all:

• All combinations yield $$s \geq 4$$, and $$s$$ must differ from $$p, q, r$$ and be $$\leq 5$$. No valid assignments.

Case $$q = 5$$: $$s = p + 10 - r \geq 1 + 10 - 4 = 7 > 5$$. No valid assignments.

Compute the probability:

Total favourable assignments: $$2 + 2 + 2 + 0 + 0 = 6$$

$$\text{Probability} = \frac{6}{120} = \frac{1}{20}$$

The correct answer is Option D: $$\dfrac{1}{20}$$.