Is this a permutations problem?

[Alex Ouroumidis]

A 100m runner is going to run in lane 3 of an 8-lane stage. If he knows that in addition to him, 3 other athletes are taking part in the race, who are placed completely randomly in the other 7 lanes, what is the probability that no opponent will run next to him (in the adjacent lanes)?
I have some ideas but don't know how to actually approach the problem.

 

[Romsek]

There are two lanes that are adjacent to lane 3. Thus the other athletes must be distributed among the remaining 5 lanes.

How many ways can you distribute 3 athletes among 5 lanes?

How many ways can you distribute 3 athletes among 7 lanes?

Thus what is the probability they seek?

 

[R.M.]

Please share your ideas. You may be on the right track (pun intended) without realizing it.

 

[pka]

A 100m runner is going to run in lane 3 of an 8-lane stage. If he knows that in addition to him, 3 other athletes are taking part in the race, who are placed completely randomly in the other 7 lanes, what is the probability that no opponent will run next to him (in the adjacent lanes)? I have some ideas but don't know how to actually approach the problem.

Although you titled this permutations, this is an occupancy problem-how are the lanes occupied? Thus we use combinations.
Lane 3 fixed, The question askes about the probability that after the random assigning of three more runners that lanes 2&4 will be un-occupied.
There are \(\dbinom{7}{3}=35\) seehere ways to assign the three. How many of those avoid lanes 2&4?

 

[Alex Ouroumidis]

Please share your ideas. You may be on the right track (pun intended) without realizing it.

Using combinations I get that there are 10 ways to distribute 3 athletes among 5 lanes and 35 ways to distribute 3 athletes among seven lanes. Does that mean that the probability I'm looking for is 10/35? Nice pun btw?

 

[Romsek]

Using combinations I get that there are 10 ways to distribute 3 athletes among 5 lanes and 35 ways to distribute 3 athletes among seven lanes. Does that mean that the probability I'm looking for is 10/35? Nice pun btw?

Yes that's correct though it can be simplified to [MATH]\dfrac{2}{7}[/MATH]