1. Total number of one-one functions

A one-one function from $\{a,b,c,d\}$ to $\{1,2,3,4,5\}$ assigns distinct values to $a,b,c,d$.

So the total number of injective functions is

2. Condition to be satisfied

We need

Let

where $x,y,z,w$ are distinct elements of $\{1,2,3,4,5\}$.

Then the condition becomes

We must count the number of ordered 4-tuples $(x,y,z,w)$ of distinct numbers from $\{1,2,3,4,5\}$ satisfying this equation.

3. Count favorable cases by choosing $y$

Since

all of $x,z,w$ must lie in $\{1,2,3,4,5\}$ and be distinct from each other and from $y$.

We check possible values of $y$.

Case 1: $y=1$

Then

Here $x,z,w$ must be distinct and chosen from $\{2,3,4,5\}$.

Check ordered pairs $(x,z)$ from $\{2,3,4,5\}$, $x\neq z$:

• If $x=2$, then $w=4-z$, giving values $1,0,-1$ for $z=3,4,5$ — not allowed.
• If $x=3$, then $w=5-z$, giving $3,1,0$ for $z=2,4,5$ — invalid/distinctness fails.
• If $x=4$, then $w=6-z$, giving $4,2,1$ for $z=2,3,5$ — invalid/distinctness fails.
• If $x=5$, then $w=7-z$, giving $5,4,3$ for $z=2,3,4$; valid cases are:
  • $(x,z,w)=(5,3,4)$
  • $(x,z,w)=(5,4,3)$

So for $y=1$, favorable cases = $2$.

Case 2: $y=2$

Then

Now $x,z,w$ are distinct and chosen from $\{1,3,4,5\}$.

Check ordered pairs $(x,z)$:

• $x=1$: $w=5-z$ gives values $2,1,0$ for $z=3,4,5$ — invalid.
• $x=3$: $w=7-z$ gives $6,3,2$ for $z=1,4,5$ — invalid.
• $x=4$: $w=8-z$ gives $7,4,3$ for $z=1,3,5$; valid: $(4,5,3)$
• $x=5$: $w=9-z$ gives $8,6,5$ for $z=1,3,4$ — invalid.

So for $y=2$, favorable cases = $1$.

Case 3: $y=3$

Then

Now $x,z,w$ are distinct and chosen from $\{1,2,4,5\}$.

Check ordered pairs:

• $x=1$: $w=7-z$ gives $5,3,2$ for $z=2,4,5$; valid: $(1,2,5)$
• $x=2$: $w=8-z$ gives $7,4,3$ for $z=1,4,5$; valid: $(2,4,4)$ not distinct, others invalid
• $x=4$: $w=10-z$ gives $9,8,5$ for $z=1,2,5$; valid: $(4,5,5)$ not distinct
• $x=5$: $w=11-z$ gives $10,9,7$ — invalid

So only one valid case:

Thus for $y=3$, favorable cases = $1$.

Case 4: $y=4$

Then

Now $x,z,w$ are distinct and chosen from $\{1,2,3,5\}$.

Check ordered pairs:

• $x=1$: $w=9-z$ gives $7,6,4$ — invalid
• $x=2$: $w=10-z$ gives $9,8,5$ — invalid except $z=5$ gives $5$ not distinct
• $x=3$: $w=11-z$ gives $10,9,8$ — invalid
• $x=5$: $w=13-z$ gives $12,11,10$ — invalid

So for $y=4$, favorable cases = $0$.

Case 5: $y=5$

Then

Now $x,z,w$ are distinct and chosen from $\{1,2,3,4\}$.

But $x+10-z\ge 7$, so $w\notin\{1,2,3,4,5\}$.

Hence favorable cases = $0$.

4. Total favorable functions

Thus total favorable cases are

5. Probability

Therefore,

So the correct option is

6. Comparison with stored answer

Stored correct answer is D: $\frac{1}{20}$.

But our careful counting gives

So I disagree with the stored answer.