Permutations: Choosing from a Single Source - without Repetition – Order Matters

Runners $$M$$ and $$N$$ are among the 5 runners in a race, and there can be no tie. How many possible results are there where $$M$$ is ahead of $$N$$?

If you want to analyze this by numbers, do the following analysis: * If $$N$$ is last, then there are {color:dark-green}4{/color} places where $$M$$ is ahead of $$N$$ (first, second, third, and fourth places), and {color:red}0{/color} places where he is behind. * If $$N$$ is in fourth place, then there are {color:dark-green}3{/color} places where $$M$$ is ahead (first, second, and third) and {color:red}1{/color} place where he is behind. * If $$N$$ is in third place, then there are exactly {color:dark-green}2{/color} places where $$M$$ is ahead and {color:red}2{/color} places where he is behind. * If $$N$$ is in second place, then there is only {color:dark-green}1{/color} place where $$M$$ is ahead and {color:red}3{/color} places where he is behind. Note that this situation is the exact symmetrical opposite of the situation where $$N$$ is in fourth place. * If $$N$$ is first, then there are {color:dark-green}0{/color} places where $$M$$ is ahead and {color:red}4{/color} places where he is behind—the exact symmetrical opposite of the situation where $$N$$ is last. If you go down the above list and sum the number of cases either way (ahead/behind), you will find that the number of places where $$M$$ is ahead ($$\color{green}{4+3+2+1+0}$$) is equal to the number of cases where $$M$$ is behind $$N$$ ($$\color{red}{0+1+2+3+4}$$). This proves that there is no real bias either way, and thus the number of possible results where $$M$$ is ahead of $$N$$ should be exactly half of the total number of results, or $$\frac{120}{2}=60$$.

Incorrect. [[snippet]]

Incorrect. [[snippet]]

Incorrect. [[snippet]]

Incorrect. [[snippet]]

Correct. [[snippet]] Considering the fact that there is no possibility of a tie, $$M$$ will be either ahead or behind $$N$$. Intuitively, there is no real reason why there would be more results with $$M$$ ahead of $$N$$ rather than with $$M$$ behind $$N$$. Thus, the number of combinations where $$M$$ is ahead of $$N$$ should be a straightforward: it is half of the total number of ways of ordering 5 runners in a race. There are $$5! = 120$$ ways of ordering 5 runners, so half of that is $$\frac{120}{2}=60$$.

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