Respuesta :
Combinations and Variations of Elements
Let's suppose we have two dogs only, A and B. They can only finish in two possible orders: AB or BA.
If we add a third dog, let's say C, the combinations (better-called variations here) are now ABC, ACB, BAC, BCA, CAB, and CBA, a total of 6 variations.
Note that we added a 3rd element and the variations changed from 2 to 6, that is, the number was multiplied by 3.
If we add a fourth dog, the total number of possible variations is 6*4 = 24
Following this very same pattern, for 9 dogs, there will be a total of
9*8*7*6*5*4*3*2 = 362880 variations.
Out of these possibilities, we are trying to find the probability that the first three places are occupied by three specific dogs, and the other 6 positions can be filled up with a random variation that will give us
6*5*4*3*2 = 720 variations.
Thus the required probability is:
[tex]\begin{gathered} p=\frac{720}{362880} \\ \text{Simplifying the result, we get:} \\ p=\frac{1}{504} \end{gathered}[/tex]