Question

How many ways can n people finish in a competition, if ties are possible? Express the answer in terms of Stirling numbers of the second kind.

Answer 1

Answer:

1. When no ties:

The quantity of stages is P(4,4)

= 4!

= 24

2. For two tie:

There are C(4,2) = 6 different ways to pick the two that tie

There are P(3,3) = 6 different ways for the "gatherings" to complete A "gathering" is either a solitary tie or the two tying By the item rule ,

there are 6*6

= 36 prospects for this case

3. For two gatherings of two tie

There are C(4,2) = 6 different ways to pick the two winning ties

The other two tie for second spot

4. For three tie with one another:

There are C(4,3) = 4 different ways to pick the two tie that tie.

There are P(2,2) = 2 different ways for the "gatherings" to wrap up By the item rule,

there are 4*2

= 8 prospects for this case

5. For each of the four tie:

There is just a single blend for this By the aggregate guideline,

the all out is

= 24 + 36 + 6 + 8 + 1 = 75

N = 75