Key Concepts and Formulas

• Permutation Formula: The number of permutations of $n$ distinct items taken $r$ at a time is given by ${}^{n}{P_{r}} = \frac{n!}{(n-r)!}$.
• Conditions for Permutations: For ${}^{n}{P_{r}}$ to be defined, the following conditions must be met:
  • $n$ must be a non-negative integer ($n \ge 0$).
  • $r$ must be a non-negative integer ($r \ge 0$).
  • $n$ must be greater than or equal to $r$ ($n \ge r$).
• Domain of a Function: The set of all possible input values ($x$) for which the function is defined.
• Range of a Function: The set of all possible output values ($f(x)$) that the function can produce.

Step-by-Step Solution

Step 1: Identify the components of the permutation in the function. The given function is $f(x) = {}^{7-x}{P_{x-3}}$. In this permutation, $n = 7-x$ and $r = x-3$.

Step 2: Determine the domain of the function by applying the conditions for permutations. We need to ensure that $n$, $r$, and the relationship between them satisfy the conditions:

• Condition 1: $n \ge 0$ and $n$ is an integer. $7-x \ge 0 \implies x \le 7$. For $7-x$ to be an integer, $x$ must be an integer (since 7 is an integer).

• Condition 2: $r \ge 0$ and $r$ is an integer. $x-3 \ge 0 \implies x \ge 3$. For $x-3$ to be an integer, $x$ must be an integer.

• Condition 3: $n \ge r$. $7-x \ge x-3$ $7+3 \ge x+x$ $10 \ge 2x$ $5 \ge x \implies x \le 5$.

Combining all the conditions:

• $x$ must be an integer.
• $x \le 7$
• $x \ge 3$
• $x \le 5$

The integers $x$ that satisfy $x \ge 3$ and $x \le 5$ are $x \in \{3, 4, 5\}$. This is the domain of the function $f(x)$.

Step 3: Evaluate the function $f(x)$ for each value in its domain.

• For $x=3$: $n = 7-3 = 4$ $r = 3-3 = 0$ $f(3) = {}^{4}{P_{0}}$ Using the formula ${}^{n}{P_{r}} = \frac{n!}{(n-r)!}$: $f(3) = \frac{4!}{(4-0)!} = \frac{4!}{4!} = 1$.

• For $x=4$: $n = 7-4 = 3$ $r = 4-3 = 1$ $f(4) = {}^{3}{P_{1}}$ Using the formula: $f(4) = \frac{3!}{(3-1)!} = \frac{3!}{2!} = \frac{3 \times 2 \times 1}{2 \times 1} = 3$.

• For $x=5$: $n = 7-5 = 2$ $r = 5-3 = 2$ $f(5) = {}^{2}{P_{2}}$ Using the formula: $f(5) = \frac{2!}{(2-2)!} = \frac{2!}{0!} = \frac{2 \times 1}{1} = 2$ (since $0! = 1$).

Step 4: Collect the output values to determine the range. The values of $f(x)$ calculated are $1$, $3$, and $2$. The range of the function is the set of these unique values: $\{1, 2, 3\}$.

Common Mistakes & Tips

• Forgetting the integer constraint: Always remember that $n$ and $r$ in permutations must be integers. This is crucial for determining the domain of $x$.
• Incorrectly applying $n \ge r$: This inequality is a fundamental requirement for permutations. Failing to apply it will lead to an incorrect domain.
• Memorizing special cases: Knowing that ${}^{n}{P_{0}} = 1$, ${}^{n}{P_{1}} = n$, and ${}^{n}{P_{n}} = n!$ can speed up calculations for specific values of $r$.

Summary To find the range of the function $f(x) = {}^{7-x}{P_{x-3}}$, we first established the domain of $x$ by ensuring that the parameters of the permutation ($n=7-x$ and $r=x-3$) met the necessary conditions: $n \ge 0$, $r \ge 0$, $n \ge r$, and that both $n$ and $r$ are integers. This led to the domain $x \in \{3, 4, 5\}$. We then evaluated $f(x)$ for each of these values, obtaining $f(3)=1$, $f(4)=3$, and $f(5)=2$. The set of these output values constitutes the range of the function.

The final answer is \boxed{{1, 2, 3}}. This corresponds to option (D).