This problem combines the concepts of one-one (injective) functions with probability and systematic enumeration. We need to find the total number of possible one-one functions and then count how many of them satisfy the given condition.

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1. Key Concepts and Formulas

   • One-one function (Injective function): A function $f: A \to B$ is one-one if every distinct element in the domain $A$ maps to a distinct element in the codomain $B$. That is, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.
   • Number of one-one functions: If set $A$ has $m$ elements and set $B$ has $n$ elements, the number of one-one functions from $A$ to $B$ is given by the permutation formula $P(n, m) = \frac{n!}{(n-m)!}$, provided $n \ge m$. If $n < m$, there are no one-one functions.
   • Probability: The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely. $P(\text{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$

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1. Step-by-Step Solution

   Step 1: Calculate the Total Number of One-One Functions

   • What we are doing: Determining the total sample space for our probability calculation.
   • Why: This forms the denominator of our probability fraction.
   • Problem Setup:
     • The domain set is $A = \{a, b, c, d\}$, so $m = |A| = 4$.
     • The codomain set is $B = \{1, 2, 3, 4, 5\}$, so $n = |B| = 5$.
   • Applying the formula: Since $n \ge m$ (i.e., $5 \ge 4$), one-one functions are possible. The total number of one-one functions from $A$ to $B$ is $P(5, 4)$. $P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5 \times 4 \times 3 \times 2 = 120$
   • Reasoning: To define a one-one function $f: \{a,b,c,d\} \to \{1,2,3,4,5\}$:
     • $f(a)$ can be any of the 5 elements in the codomain.
     • Since $f$ must be one-one, $f(b)$ must be different from $f(a)$, leaving 4 choices.
     • Similarly, $f(c)$ must be different from $f(a)$ and $f(b)$, leaving 3 choices.
     • Finally, $f(d)$ must be different from $f(a)$, $f(b)$, and $f(c)$, leaving 2 choices. Thus, the total number of ways to assign distinct values from $B$ to the elements of $A$ is $5 \times 4 \times 3 \times 2 = 120$.

   Step 2: Analyze the Given Condition and Set up for Favorable Cases

   • What we are doing: Understanding the constraint for favorable outcomes and preparing for systematic enumeration.
   • Why: This condition defines which of the 120 functions are "favorable".
   • The Condition: $f(a) + 2f(b) - f(c) = f(d)$.
   • Notation: Let $f(a) = x_1$, $f(b) = x_2$, $f(c) = x_3$, and $f(d) = x_4$.
   • Constraints on $x_i$: Since $f$ is a one-one function, $x_1, x_2, x_3, x_4$ must be distinct values chosen from the set $\{1, 2, 3, 4, 5\}$.
   • Rewriting the equation: $x_1 + 2x_2 - x_3 = x_4$ Rearranging for easier analysis: $x_1 + 2x_2 = x_3 + x_4$
   • Strategy for finding favorable cases: We will systematically iterate through possible values for $x_2$ because its coefficient of 2 makes it a good starting point for constraining the sums. For each choice of $x_2$ and $x_1$, we will determine the required sum $x_3+x_4$ and find distinct pairs from the remaining available numbers.
   • Initial Constraints on $x_2$:
     • $x_2$ must be from $\{1, 2, 3, 4, 5\}$.
     • Consider the maximum value for $x_2$: If $x_2 = 5$, then $x_1, x_3, x_4$ must be distinct values chosen from $\{1, 2, 3, 4\}$. The equation becomes $x_1 + 2(5) = x_3 + x_4 \implies x_1 + 10 = x_3 + x_4$. The smallest possible value for $x_1$ (from $\{1,2,3,4\}$) is 1, so $x_1+10 \ge 1+10=11$. The largest possible sum for $x_3+x_4$ (distinct and from $\{1,2,3,4\}$) is $4+3=7$. Since $11 > 7$, there are no solutions when $x_2=5$.
     • Thus, $x_2$ can only be $1, 2, 3, 4$.

   Step 3: Systematically Find the Number of Favorable Functions

   • What we are doing: Enumerating all possible quadruplets $(x_1, x_2, x_3, x_4)$ that satisfy the condition and distinctness.
   • Why: This gives us the numerator for the probability calculation.

   We check each possible value for $x_2$:

   • Case 1: $x_2 = 1$

     • The equation becomes $x_1 + 2(1) = x_3 + x_4 \implies x_1 + 2 = x_3 + x_4$.
     • The values $x_1, x_3, x_4$ must be distinct and chosen from $\{2, 3, 4, 5\}$ (since $x_2=1$ is used).
     • Iterate $x_1$ from the available set:
       • If $x_1 = 2$: $x_3 + x_4 = 4$. Remaining for $x_3, x_4$: $\{3,4,5\}$. Minimum sum $3+4=7$. No solution.
       • If $x_1 = 3$: $x_3 + x_4 = 5$. Remaining for $x_3, x_4$: $\{2,4,5\}$. Minimum sum $2+4=6$. No solution.
       • If $x_1 = 4$: $x_3 + x_4 = 6$. Remaining for $x_3, x_4$: $\{2,3,5\}$. Possible sums: $2+3=5, 2+5=7, 3+5=8$. No solution.
       • If $x_1 = 5$: $x_3 + x_4 = 7$. Remaining for $x_3, x_4$: $\{2,3,4\}$. The distinct pairs summing to 7 are $(3,4)$ and $(4,3)$.
         • This gives two solutions: $(x_1, x_2, x_3, x_4) = (5, 1, 3, 4)$ and $(5, 1, 4, 3)$.
         • (2 favorable functions)

   • Case 2: $x_2 = 2$

     • The equation becomes $x_1 + 2(2) = x_3 + x_4 \implies x_1 + 4 = x_3 + x_4$.
     • The values $x_1, x_3, x_4$ must be distinct and chosen from $\{1, 3, 4, 5\}$ (since $x_2=2$ is used).
     • Iterate $x_1$ from the available set:
       • If $x_1 = 1$: $x_3 + x_4 = 5$. Remaining for $x_3, x_4$: $\{3,4,5\}$. Minimum sum $3+4=7$. No solution.
       • If $x_1 = 3$: $x_3 + x_4 = 7$. Remaining for $x_3, x_4$: $\{1,4,5\}$. Possible sums: $1+4=5, 1+5=6, 4+5=9$. No solution.
       • If $x_1 = 4$: $x_3 + x_4 = 8$. Remaining for $x_3, x_4$: $\{1,3,5\}$. The distinct pairs summing to 8 are $(3,5)$ and $(5,3)$.
         • This gives two solutions: $(x_1, x_2, x_3, x_4) = (4, 2, 3, 5)$ and $(4, 2, 5, 3)$.
         • (2 favorable functions)
       • If $x_1 = 5$: $x_3 + x_4 = 9$. Remaining for $x_3, x_4$: $\{1,3,4\}$. Possible sums: $1+3=4, 1+4=5, 3+4=7$. No solution.

   • Case 3: $x_2 = 3$

     • The equation becomes $x_1 + 2(3) = x_3 + x_4 \implies x_1 + 6 = x_3 + x_4$.
     • The values $x_1, x_3, x_4$ must be distinct and chosen from $\{1, 2, 4, 5\}$ (since $x_2=3$ is used).
     • Iterate $x_1$ from the available set:
       • If $x_1 = 1$: $x_3 + x_4 = 7$. Remaining for $x_3, x_4$: $\{2,4,5\}$. The distinct pairs summing to 7 are $(2,5)$ and $(5,2)$.
         • This gives two solutions: $(x_1, x_2, x_3, x_4) = (1, 3, 2, 5)$ and $(1, 3, 5, 2)$.
         • (2 favorable functions)
       • If $x_1 = 2$: $x_3 + x_4 = 8$. Remaining for $x_3, x_4$: $\{1,4,5\}$. Possible sums: $1+4=5, 1+5=6, 4+5=9$. No solution.
       • If $x_1 = 4$: $x_3 + x_4 = 10$. Remaining for $x_3, x_4$: $\{1,2,5\}$. Maximum sum $2+5=7$. No solution.
       • If $x_1 = 5$: $x_3 + x_4 = 11$. Remaining for $x_3, x_4$: $\{1,2,4\}$. Maximum sum $2+4=6$. No solution.

   • Case 4: $x_2 = 4$

     • The equation becomes $x_1 + 2(4) = x_3 + x_4 \implies x_1 + 8 = x_3 + x_4$.
     • The values $x_1, x_3, x_4$ must be distinct and chosen from $\{1, 2, 3, 5\}$ (since $x_2=4$ is used).
     • Iterate $x_1$ from the available set:
       • If $x_1 = 1$: $x_3 + x_4 = 9$. Remaining for $x_3, x_4$: $\{2,3,5\}$. Maximum sum $3+5=8$. No solution.
       • If $x_1 = 2$: $x_3 + x_4 = 10$. Remaining for $x_3, x_4$: $\{1,3,5\}$. Maximum sum $3+5=8$. No solution.
       • If $x_1 = 3$: $x_3 + x_4 = 11$. Remaining for $x_3, x_4$: $\{1,2,5\}$. Maximum sum $2+5=7$. No solution.
       • If $x_1 = 5$: $x_3 + x_4 = 13$. Remaining for $x_3, x_4$: $\{1,2,3\}$. Maximum sum $2+3=5$. No solution.

   • Total Number of Favorable Functions: Summing up the solutions from each case: $2 + 2 + 2 = 6$. There are 6 one-one functions that satisfy the given condition.

   Step 4: Calculate the Probability

   • What we are doing: Combining the results from Step 1 and Step 3.
   • Why: This is the final answer to the question.
   • Probability = (Number of Favorable Functions) / (Total Number of One-One Functions) $P = \frac{6}{120} = \frac{1}{20}$

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1. Common Mistakes & Tips

   • Distinctness is Paramount: Always remember that for one-one functions, $f(a), f(b), f(c), f(d)$ must all be different numbers. This is a crucial constraint for both calculating total functions and favorable ones. Ensure that chosen $x_1, x_2, x_3, x_4$ are distinct from each other and from the already assigned values.
   • Systematic Enumeration: When dealing with conditions involving sums or equations, a systematic approach (like iterating through one variable and checking possibilities for others) is essential. This prevents missing cases or accidental double-counting.
   • Range and Bounds Analysis: Before detailed enumeration, quickly check the maximum and minimum possible values of sums or expressions to eliminate large branches of impossible cases (e.g., the $x_2=5$ case in this problem). This saves significant time.

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1. Summary

   This problem required a two-pronged approach: first, calculating the total number of one-one functions from a set of 4 elements to a set of 5 elements using permutations. Second, systematically enumerating the specific one-one functions that satisfy the given algebraic condition $f(a) + 2f(b) - f(c) = f(d)$, ensuring all values $f(a), f(b), f(c), f(d)$ are distinct and within the codomain. By carefully breaking down the enumeration into cases based on $f(b)$ and then $f(a)$, we identified 6 such favorable functions. The final probability is the ratio of favorable functions to the total number of functions.

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1. Final Answer

The final answer is $\boxed{{1 \over {20}}}$, which corresponds to option (D).