Runners and their chance of winning Six runners are entered in a track meet, and have equal ability. What is the probability that
a) they will finish in ascending order of their ages? 
b) Shanaze will finish first or Tanya will finish second?
c) Shanaze and Tanya will not finish back-to-back?
 A: $A) \frac 1{6!}$Only one possible way for them to finish in right order out of all of the possible ways.
$B)$ Chance of Shanaze finishing first is $\frac 16$, and chance of Tanya finishing second is $\frac 16$, subtract the probability that both events will happen from the combined probability of the two: $\frac 16+\frac 16-\frac 1{36}=\frac {11}{36}.$
$C)$ Chance of Shanaze and Tanya not finishing back-to back is $\frac 23$
A: Part a:
There are $6!$ possible arrangements, one of which is the one you want, so the probability is 
$$
\frac{1}{6!}
$$
Part b: By inclusion exclusion
$$
P(A\cup B)=P(A)+P(B)-P(A\cap B)=1/6+1/6-1/36=\frac{11}{36}
$$
For part c: Pretend the two people with names are holding hands, how many possible arrangements are there? 
$$
2*5*4!
$$
All the spots the pair can occupy, corrected by the number of ways to swap a pair, times the number of ways to permute the rest. So we take the complement of this and divide by the total arrangements as in part a to get
$$
\frac{6!-240}{6!}=\frac{2}{3}
$$
