
Number of Ways to Create An Ordered List or a Group

 These are counting problems. They are used here as an intro to important ideas
in probability. They should be considered before reading the page Binomial Distribution.

 1. A club has 4 members and wishes to elect a president, vice president, secretary, and treasurer. In how many ways can this be done?
 There are 4 ways to select a president, but then only 3 ways to choose the vp, but then only 2 ways to choose the secretary, and
then only 1 way to choose the treasurer,

 so, there are 4(3)(2)(1) or 24 ways to do this.

 Using members A, B, C, and D, here is a tree diagram and the sample space with 24 ways.


 2. A club has n members and wishes each member to hold an elected office, beginning with the president, vice president, secretary, and treasurer. In how many ways can this be done?
 The answer is n factorial ways. There are n ways to fill the first office, n1 ways to fill the 2nd office, n2 ways to
fill the 3rd office, and so on. There are n! or n(n1)(n2)(n3) ... (3)(2)(1) ways.

 3. A club has 4 members and wishes to elect a president and then a vice president. In how many ways can this be done?
 There are (4)(3) or 12 ways to do this.

 Here order counts. A list is required. For a source set of n elements, if order is important and x elements are
chosen for the ordered list, a permutation of n things taken x at a time is required, P_{n,x}. There are n! / (n x)! ways to do this.

 
 n!   (number of ordered lists given n things) 
P_{n,x} = 
 = 

 (nx)!   (shorten the list so only x things are used) 

 In this problem, an ordered list of 4 elements, take 2 elements is required, P_{4,2}.
There are 4! / (42)! = 24/2 = 12 ways to do this.
 4! 
 4(3)(2)(1) 

P_{4,2} = 

= 

= 12 
 (42)! 
 (2)(1) 


 4. A club has 4 members and wishes to form a committee of 2 member. In how many ways can this be done?
 This time order does not count, so, AB is really BA and should not be counted as different committee.
 This time a combination is used. A combination is a permutation divided to remove duplicates.



 n!   (number of ordered lists given n things) 
C_{n,x} = 
 = 

 (nx)! x! 
 (shorten to only x things)(remove duplicates) 

 4!   (4)(3)(2)(1) 

C_{4,2} = 
 = 
 = 6 
 (42)! 2!   (2)(1)(2)(1) 

 5. A club has n members and wishes to form a committee of r member. In how many ways can this be done?
 Order does not count. Use a combination.


