This is from Discrete mathematics and its application by Kenneth H Rosen
The signless Stirling number of the first kind $c(n,k)$ where $k$ and $n$ are integers with $1\leq k\leq n$, equals the number of ways to arrange $n$ people around $k$ circular tables with at least one person seated at each table, where two seatings of m people around a circular table are considered same if everyone has the same left neighbor and the same right neighbor. Then prove $\sum_{j=1}^nc(n,j)=n!$ if $n$ is a positive integer
I know $c(n,n)=1$,$c(n,n-1)=n_{C_2}$,$c(n,1)=n-1!$ But I don't know how to prove the summation.
@GerryMyerson i solved it with induction. Please check if it is correct
Base case: n=1 $\sum_{j=1}^1c(n,j)=1!$ which holds true because there is only one way to sit one person on one circular table.
Let P(n) holds. Now there is recurrence relation followed by c(n,k), which is $c(n + 1, k) = c(n, k − 1) + nc(n, k)$
Now for P(n+1) $\sum_{j=1}^{n+1}c(n+1,j)$=$\sum_{j=1}^{n+1}(c(n,j-1)+n*c(n,j))$
=$\sum_{j=1}^{n+1}c(n,j-1)$+$n*\sum_{j=1}^{n+1}c(n,j)$
Now let j-1=i so first sum becomes $\sum_{i=0}^{n}c(n,i)$+$n*\sum_{j=1}^{n}c(n,j)+n*c(n,n+1)$ now when i=0 c(n,0)=0 and also c(n,n+1)=0. So
$\sum_{i=1}^{n}c(n,i)$+n*n! because P(n) is true.
=n!+n*n!=n+1!. So P(n+1) is also true. Hence proved