How do you define P(n,0)=1 theoretically?

Question

Well we can define C(n,0)=1 like "Selecting nothing is like selecting everything which can be done in only one way" I'm not sure of P(n,0) tho

Answer 1

You are saying the main reason we define $C(n,0)=1$ is because $C(n,n)=1$, and the symmetry $C(n,k)=C(n,n-k)$. However, there is a more direct reason. $C(n,k)$ is defined as the number of subsets of size $k$ in an $n$-element set. Therefore, $C(n,0)$ should be the number of $0$-element subsets. Every set has exactly one zero element subset, the empty set, so $C(n,0)=1$.

The same logic can be used to define $P(n,0)$. We define $P(n,k)$ as the number of lists of length $k$ whose entries are distinct elements of an $n$-element set. Therefore, $P(n,0)$ would be the number of lists of length $0$. There is exactly one list of length zero; the empty list. Therefore, $P(n,0)=1$.

The concept of an "empty list" might require more explanation. Here are a couple of viewpoints.

• A list of length $n$ is a function from the set of first $n$ integers to the set of potential entries of the list. When $n=0$, it is a function defined on the empty set. There is exactly one function defined on the empty set; to define a function, you need to define its output for each element of its domain, but if the domain is the empty set, there is nothing to define.

• A list of length $3$ looks like $(a,b,c)$. A list of length $1$ looks like $(a)$. A list of length $0$ looks like $()$. There are no entries, so there is only one possible empty list.

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Alternatively, using the identity $$ P(n+1,k+1)=(n+1)\cdot P(n,k) $$ with $k=0$, you get $P(n+1,1)=(n+1)P(n,0)$. On the other hand, you obviously have $P(n+1,1)=n+1$, so $n+1=(n+1)P(n,0),$ so $P(n,0)=1$.