Question

in the final event of a track meet, 6 runners run a 100 meter dash. A) How many possible arrangements are there for the medal winners (gold, silver and bronze)? B) state whether it is a combination or permutation

Answer 1

Answer:

120 possible arrangements.

It is a permutation

Step-by-step explanation:

Imagine that you are going to give the gold medal to any of the 6 runners. There would be 6 possibilities; now, imagine that you are going to give the gold and the silver medals random in the same way: you could give the gold medal to a person and you would have 5 possibilities to give the silver medal (you can't give both medals to the same person). So, note that for each possibility to give the gold medal there would be 5 possibilities to give the silver one.

Analogously, if you give the three medals random and supposing that you have already given the gold one, for each possibility to give the silver medal you would have 4 possibilities to give the bronze one. That is the reason for the calculation to be a multiplication:

N= Total number possibilities:

$N=6*5*4=120$

This also can be seen as a permutation. A permutation is a reorganization of elements where the order matters. In this case, it is not just relevant who are the winners (who receive the medal), but their order. A combination just helps us to make groups without taking in mind the order, but the permutation does consider that.

The permutation can be learnt by the formula:

$N=\frac{p!}{(p-n)!}$

where p is the total ef elements and n is the amount of elements of each subgroup that I want to built. In this case, our population 'p'=6, and we want to organize them in groups of 3 (n=3) where it is important for us the order:

$N=\frac{6!}{(6-3)!}=6*5*4=120$

Answer 2

6 runners, 3 medals, how many ways to give them? well, runner say 2 could get gold or not, bronze or not and silver or not, and so can any other runnner, and the order "does matter", it makes a distinction, thus, is a Permutation

$\bf _nP_r=\cfrac{n!}{(n-r)!}\qquad \qquad _6P_3=\cfrac{6!}{(6-3)!}$