The probability that a randomly chosen one-one function from the set {a, b, c, d} to the set {1, 2,

Q: Why did we use permutations ($^5P_4$) instead of combinations?

Great question! A function maps specific inputs to specific outputs. If $f(a)=1$ and $f(b)=2$, that is a different function than $f(a)=2$ and $f(b)=1$. Because the order of assignment matters, we must use permutations.

Q: Why can't we just pick random numbers for $f(a), f(b), f(c), f(d)$?

Because the function is 'one-one.' If you pick $f(a)=1$ and $f(b)=1$, you've violated the definition of a one-one function. Every element in the domain must map to a unique element in the codomain.

Q: How do we know we didn't miss any cases?

By anchoring the variable with the largest coefficient ($f(b)$), we created a systematic hierarchy. Once $f(b)$ becomes too large, the left side of the equation grows faster than the right side can possibly match, proving no further solutions exist.

Question

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 :

Options:

(A) $\frac{1}{24}$

(B) $\frac{1}{40}$

(C) $\frac{1}{30}$

(D) $\frac{1}{20}$

Sigma Percentile · Accepted Answer

### Analyzing the Landscape

We begin by defining our playground. We have a domain, Set $A = \{a, b, c, d\}$, and a codomain, Set $B = \{1, 2, 3, 4, 5\}$.

We are constructing a one-one function $f: A 	o B$. The term 'one-one' is our golden rule: every element in $A$ must map to a unique element in $B$. No two elements in $A$ can share the same destination.

### The Sample Space

Before we hunt for the specific condition, we must know the total size of our universe. How many one-one functions exist from $A$ to $B$?

We have 5 choices for $f(a)$, 4 for $f(b)$, 3 for $f(c)$, and 2 for $f(d)$. Mathematically, this is the number of permutations of 5 items taken 4 at a time:

$$n(S) = {}^5P_4 = 5 	imes 4 	imes 3 	imes 2 = 120$$

This value represents our denominator.

### The Constraint

Now, we face the heart of the problem: $f(a) + 2f(b) - f(c) = f(d)$. This looks intimidating, but let us simplify it.

By rearranging, we get:

$$f(a) + 2f(b) = f(c) + f(d)$$

We need to find how many assignments of values from $\{1, 2, 3, 4, 5\}$ satisfy this equality while respecting the one-one constraint.

### The Strategy of Anchoring

Why did we choose to focus on $f(b)$? Because it is multiplied by 2; it is the 'heavyweight' of the equation. By fixing $f(b)$ first, we drastically reduce the number of cases we need to check.

**Case 1: $f(b) = 1$**
The equation becomes $f(a) + 2 = f(c) + f(d)$. We must pick $f(a), f(c), f(d)$ from the remaining set $\{2, 3, 4, 5\}$. 
If we set $f(a) = 5$, then $5 + 2 = 7$. We need $f(c) + f(d) = 7$ using the remaining set $\{2, 3, 4\}$. The only pair that sums to 7 is $\{3, 4\}$. Since $f(c)$ and $f(d)$ can be swapped, this gives us **2** valid functions.

**Case 2: $f(b) = 2$**
The equation becomes $f(a) + 4 = f(c) + f(d)$. Using the remaining set $\{1, 3, 4, 5\}$, if we set $f(a) = 4$, then $4 + 4 = 8$. We need $f(c) + f(d) = 8$ using $\{1, 3, 5\}$. The pair $\{3, 5\}$ works. Again, swapping gives us **2** valid functions.

**Case 3: $f(b) = 3$**
The equation becomes $f(a) + 6 = f(c) + f(d)$. Using the remaining set $\{1, 2, 4, 5\}$, if we set $f(a) = 1$, then $1 + 6 = 7$. We need $f(c) + f(d) = 7$ using $\{2, 4, 5\}$. The pair $\{2, 5\}$ works. This gives us **2** more valid functions.

### The Boundary

What happens if $f(b) = 4$? The left side becomes $f(a) + 8$. Even with the smallest possible $f(a) = 1$, the left side is 9.

The maximum possible sum for $f(c) + f(d)$ using the remaining numbers is $5 + 3 = 8$. Since $9 > 8$, there are no solutions here. The same logic applies to $f(b) = 5$.

### Final Calculation

We have systematically exhausted the possibilities. We found $2 + 2 + 2 = 6$ favorable cases.

The probability is the number of favorable cases divided by the total sample space:

$$P = \frac{6}{120} = \frac{1}{20}$$

The final probability is **$\frac{1}{20}$**.

Animated Solution for Mathematics - Probability: 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 $f (a) + 2 f (b) - f (c) = f (d)$ is :

Select Answer:

Visualized Solution

Defining the Sets

One-One Function Condition

Total Sample Space

The Given Condition

Rearranging the Equation

Case 1: $f (b) = 1$

Solving Case 1

Case 2: $f (b) = 2$

Solving Case 2

Case 3: $f (b) = 3$

Solving Case 3

Checking $f (b) \geq 4$

Final Probability

The Sigma Insight: Classical Definition of Probability

Analyzing the Landscape

The Sample Space

The Constraint

The Strategy of Anchoring

The Boundary

Final Calculation

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