Permutations

[CPerry]

Apologies if this is the wrong area but I have a problem about permutations.

There is a 6 runner race happening and I want to know how many permutations there are of three specific runners finishing this race. I am hoping somebody could explain their working out and if there is a neat tidy little formula which can be used. I can work it out for two runners such as the following:

(1st+2nd),(1st+3rd),(1st+4th),(1st+5th),(1st+6th)
(2nd+1st),(2nd+3rd),(2nd+4th),(2nd+5th),(2nd+6th)
(3rd+1st),(3rd+2nd),(3rd+4th),(3rd+5th),(3rd+6th)
(4th+1st),(4th+2nd),(4th+3rd),(4th+5th),(4th+6th)
(5th+1st),(5th+2nd),(5th+3rd),(5th+4th),(5th+6th)
(6th+1st),(6th+2nd),(6th+3rd),(6th+4th),(6th+5th)

There are 30 possible permutations here, what is the formula if there were 3 runners I cared about or 4 or 5 or 6? What about if there were 10 runners in the race altogether? Any formulae, explanations, guidance or direction would be massively appreciated. Thank you!

 

[Harry_the_cat]

The 30 you obtained (the hard way) above can more easily be found by considering the number of positions that the two runners can take when finishing the race.
Runner A could finish in 6 positions, Runner B could finish in 5 positions (since Runner A has already claimed one position which can't be repeated). Multiplying 6x5 gives 30.
Can you extend this idea to three runners?

 

[Steven G]

Apologies if this is the wrong area but I have a problem about permutations.

There is a 6 runner race happening and I want to know how many permutations there are of three specific runners finishing this race. I am hoping somebody could explain their working out and if there is a neat tidy little formula which can be used. I can work it out for two runners such as the following:

(1st+2nd),(1st+3rd),(1st+4th),(1st+5th),(1st+6th)
(2nd+1st),(2nd+3rd),(2nd+4th),(2nd+5th),(2nd+6th)
(3rd+1st),(3rd+2nd),(3rd+4th),(3rd+5th),(3rd+6th)
(4th+1st),(4th+2nd),(4th+3rd),(4th+5th),(4th+6th)
(5th+1st),(5th+2nd),(5th+3rd),(5th+4th),(5th+6th)
(6th+1st),(6th+2nd),(6th+3rd),(6th+4th),(6th+5th)

There are 30 possible permutations here, what is the formula if there were 3 runners I cared about or 4 or 5 or 6? What about if there were 10 runners in the race altogether? Any formulae, explanations, guidance or direction would be massively appreciated. Thank you!

Have you noticed that each line you wrote has 5 entries? Why is that?? Think. There are 6 lines? Why is that? Can you get 30 from 5 and 6? How?
Now what would you do to each of your 30 entries to include the runner who came in 3rd place?
I will help you with one of them, the bold one above (4th + 3rd) which I will write as (4,3)
So you can have (4,3,1), (4,3,2), (4,3,5) and (4,3,6). There are 4 of them. Why 4? Think about that! Try to modify others to have the 1st, 2nd and 3rd place winners. Will you always have to add 4 for each of your 30 listed above? Why? Do now you see the answer that pka arrived at (if not directly from that post)?

 

[pka]

There is a 6 runner race happening and I want to know how many permutations there are of three specific runners finishing this race. I am hoping somebody could explain their working out and if there is a neat tidy little formula which can be used. I can work it out for two runners such as the following:
(1st+2nd),(1st+3rd),(1st+4th),(1st+5th),(1st+6th)
There are 30 possible permutations here, what is the formula if there were 3 runners I cared about or 4 or 5 or 6? What about if there were 10 runners in the race altogether? Any formulae, explanations, guidance or direction would be massively appreciated.

Honestly I have no idea what you mean. But having read the other responses, here is my response.
\(\displaystyle \mathcal{P}_j^N=\frac{N!}{(N-j)!}\) is the number of permutations of \(\displaystyle N\) taken \(\displaystyle j\) at time.
Thus \(\displaystyle \mathcal{P}_2^6=\frac{6!}{(6-2)!}=30\) as you ready know. BUT \(\displaystyle \mathcal{P}_3^6=\frac{6!}{(6-3)!}=6\cdot5\cdot4=120\)
So now you have your 'short formula'.

 

[CPerry]

Amazing, thank you everybody! That fits my model perfectly. Cheers for such detailed responses, was just one of those silly mind blanks.