Ahorse race has 13 entries. assuming that there are no ties, what is the probability that the four horses owned by one person finish first, second, third and fourth?

Well, there are 13 places that each horse can place. That means the first place has 13 options, second place has 12 options, third place has 11 options, etc. Though, we only care up to fourth place. When doing probability, we're multiplying in this case. So, here's the process:

How many ways all the horses can place = 13! = 6227020800
How many ways the first four places can place = 13*12*11*10 = 17160

Divide the second by the first and multiply by 100 for the probability as a percent:

(17160/6227020800)*100% ≈ 0.00027557319%

The solution would be like this for this specific problem:

4/13 * 3/12* 2/11 * 1/10 = 0.0014

So the probability that the four horses owned by one person finish first, second, third and fourth is 0.0014. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

Then just do it (lol)

(m+11n)^2=(m+11n)(m+11n)=m^2+11mn+11mn+121n^2=m^2+22mn×121n^2 for example; use foil: (a+b)(a+b)first. a^2outside. abinside. ablast. b^2