Counting permutations and subsets

Question

$P(5,7)$

My answer: $(7! / (7-5)!)\cdot (7! / 2!) = 2520$.

However some people told me I was wrong and the answer should be $0$.

Answer 1

No, you've confused it with $P(7,5)=\dfrac{7!}{2!}$.

Where as $P(5,7)=\dfrac{5!}{(-2)!}=0$

Answer 2

Your friends are correct. It is not possible to select $7$ elements from a set with only five elements. Therefore, $P(5, 7) = 0$.

We could also apply the Multiplication Principle. Since the set contains five elements, there are five ways to make the first selection. Since selections are made without replacement, there are four ways to select the second element, three ways to select the third element, two ways to select the fourth element, one way to select the fifth element, zero ways to select the nonexistent sixth element, and zero ways to select the nonexistent seventh element. Hence, $$P(5, 7) = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 0 \cdot 0 = 0$$

Let $n, k$ be nonnegative integers. The number of ordered selections of $k$ elements from a set with $n$ elements is $$P(n, k) = n(n - 1)(n - 2) \cdots [n - (k - 1)] = n(n - 1)(n - 2) \cdots (n - k + 1)$$ Notice that if $k > n$, then one of the terms in this product is $n - n = 0$. Otherwise, we can multiply and divide by $(n - k)!$ to obtain \begin{align*} P(n, k) & = n(n - 1)(n - 2) \cdots (n - k + 1)\\ & = n(n - 1)(n - 2) \cdots (n - k + 1) \cdot \frac{(n - k)!}{(n - k)!}\\ & = \frac{n!}{(n - k)!} \end{align*} which yields the formula
$$P(n, k) = \begin{cases} \dfrac{n!}{(n - k)!} & \text{if $k \leq n$}\\ 0 & \text{if $k > n$} \end{cases} $$