Question

In how many ways can a five horse race end, allowing for the possibility that

two horses tie?

Answer 1

LETS FIRST SOLVE FOR 4 HORSES

1.No ties:

The number of permutations is P(4,4) = 4! = 24

2.Two horses tie:

There are C(4,2) = 6 ways to choose the two horses that tie There are P(3,3) = 6 ways for the “groups” to finish A “group” is either a single horse or the two tying horses By the product rule, there are 6*6 = 36 possibilities for this case

3.Two groups of two horses tie

There are C(4,2) = 6 ways to choose the two winning horses The other two horses tie for second place

4.Three horses tie with each other:

There are C(4,3) = 4 ways to choose the two horses that tie There are P(2,2) = 2 ways for the “groups” to finish By the product rule, there are 4*2 = 8 possibilities for this case

5.All four horses tie:

There is only one combination for this By the sum rule, the total is 24+36+6+8+1 = 75

NOW,  The answers for 0, 1, 2, 3, or 4 horses are 1, 1, 3, 13, and 75

there are five horses. If just one horse is in first place, there are (5C1) ways to choose that horse, and 75 ways to order the remaining four horses. If there are two horses in first place, there are (5C2) ways to choose them, and 13 ways to order the rest. And so on. Thus the answer is

(5C1)⋅75+(5C2)⋅13+(5C3)⋅3+(5C4)⋅1+(5C5)⋅1=541