Answer:
a) see below for a listing of possible orders
b) 0.6
Explanation:
You are asked to list the possible permutations of 5 runners, taken 3 at a time. Then, you are asked for the fraction of those that have Andrew among them.
Permutations
The formula for the number of permutations of n objects taken k at a time is ...
P(n, k) = n!/(n -k)!
Then the number of ways three can place from five contestants is ...
P(5, 3) = 5!/(5 -3)! = 5·4·3 = 60
Listing
For listing purposes, we can represent the runners with the letters ...
- A: Andrew
- B: Bill
- C: Francis
- D: Asif
- E: Sean
The order we choose for listing the permutations is to list a combination of three finishers, followed by the other 5 orders in which they can finish. In order of first, second, third, we can have ...
ABC ACB BAC BCA CAB CBA ABD ADB BAD BDA DAB DBA
ABE AEB BAE BEA EAB EBA ACD ADC CAD CDA DAC DCA
ACE AEC CAE CEA EAC ECA ADE AED DAE DEA EAD EDA
BCD BDC CBD CDB DBC DCB BCE BEC CBE CEB EBC ECB
BDE BED DBE DEB EBD EDB CDE CED DCE DEC ECD EDC
P(Andrew shows)
In the above list, the first three of the five lines all have Andrew as one of those who "show" in the race.
P(Andrew shows) = 3/5 = 0.6