Interpret the permutation expression

Given $f(x) = {}^{7-x}P_{x-3}$ and using ${}^nP_r = \frac{n!}{(n-r)!}$ we get $f(x) = \frac{(7-x)!}{(7-x-(x-3))!} = \frac{(7-x)!}{(10-2x)!}.$

Find the domain from permutation conditions

For ${}^nP_r$ to be defined in elementary combinatorics, we need:

• $n \ge 0$
• $r \ge 0$
• $r \le n$

Here, $n = 7-x, \qquad r = x-3.$

So the conditions become:

• $7-x \ge 0 \Rightarrow x \le 7$
• $x-3 \ge 0 \Rightarrow x \ge 3$
• $x-3 \le 7-x \Rightarrow 2x \le 10 \Rightarrow x \le 5$

Combining all: $3 \le x \le 5.$

Since permutation parameters are integers, $x$ must be an integer. Hence $x \in \{3,4,5\}.$

Compute the function values

• For $x=3$: $f(3) = {}^{4}P_{0} = 1$

• For $x=4$: $f(4) = {}^{3}P_{1} = 3$

• For $x=5$: $f(5) = {}^{2}P_{2} = 2! = 2$

Write the range

Therefore, the set of all possible values is $\{1,2,3\}.$

Match with options

This corresponds to Option D.

Compare with stored correct answer

Stored correct answer: D

Our derived answer: D

Hence, they agree.