Common Combination Counts
n-ball m-box assignment problem
The ball is different, the box is different, the box can be empty: m^n Each student has m choices
The ball is the same, the box is different, and the box cannot be empty: C(n-1,m-1) clapboard method
The ball is the same, the box is different, and the box can be empty: C(n+m-1,m-1) put one for each box first, and then use the partition method
Ring arrangement: (n-1)! There are n! kinds of linear arrangements, and each ring arrangement contains n kinds of linear arrangements, so divide n
Stirling numbers of the first kind: The number of ways to divide a set of n elements into k ring arrangements: S1(n+1,k) = S1(n,k-1) + n*S2(n,k) The first One is to form a ring by itself, and the second is to put this element to the left of any element
Stirling numbers of the second kind: the number of ways to divide a set of n elements into k non-empty sets S2(n,k) = S2(n-1,k-1) + k*S2(n-1,k) The first is to form a set by itself, and the second is to put this element into any set