This problem asks us to evaluate the permutation $^{15}P_4$.

The formula for permutations is:

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

Here, $n = 15$ and $r = 4$. Substituting these values into the formula:

$^{15}P_4 = \frac{15!}{(15-4)!} = \frac{15!}{11!}$

To simplify, we only calculate the product of the first four terms of $15!$ (as the $11!$ in the denominator cancels out the remaining factorial terms):

$^{15}P_4 = 15 \cdot 14 \cdot 13 \cdot 12$

Now, calculate step by step: $15 \cdot 14 = 210$ $210 \cdot 13 = 2730$ $2730 \cdot 12 = 32760$

Thus: $^{15}P_4 = 32,760$

Final Answer:

$^{15}P_4 = 32,760$

Would you like a deeper explanation or additional examples?

Related Questions:

1. What is the difference between permutations and combinations?
2. How do you calculate permutations when $r = n$?
3. Can you explain the significance of $n!$ in the formula for permutations?
4. How does $^{n}P_r$ change as $r$ increases?
5. What is $^{15}C_4$ (the combination counterpart to $^{15}P_4$)?

Tip:

Remember, permutations consider the order of arrangement, whereas combinations do not!