Math. Permutation. _r = n! / (n-r)! = 7! / (7-3)!

The formula for permutations is given by:

$$^nP_r = \frac{n!}{(n-r)!}$$

In this problem, $n=7$ and $r=3$.

Substitute $n=7$ and $r=3$ into the formula:

$$^7P_3 = \frac{7!}{(7-3)!}$$

Subtract the numbers in the parenthesis:

$$^7P_3 = \frac{7!}{4!}$$

Recall that $n! = n \times (n-1) \times (n-2) \times \dots \times 1$.
Expand $7!$ until $4!$ to simplify the expression:

$$^7P_3 = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} = \frac{7 \times 6 \times 5 \times 4!}{4!}$$

Cancel $4!$ from the numerator and the denominator:

$$^7P_3 = 7 \times 6 \times 5$$ $$^7P_3 = 42 \times 5$$ $$^7P_3 = 210$$

The final answer is $\text{210}$.