Section 5.2 Problem 6
In how many ways can a club with 17 members elect a president (P), vice president (VP), and secretary (assuming no person can fill more than one office) under the following restriction:
Part a) No restrictions
Use the following selection procedure:
Elect a vice president (17 possibilities)
Elect a president (16 possibilities)
Elect a secretary (15 possibilities)
Total: 17\cdot 16\cdot 15 choices total
Part c) Sam will only serve as president if Mary is named vice president.
Solution 1: Direct
Use the following selection procedure:
- Elect a VP. Is that VP Mary?
Yes (1 possibility)
Elect a president. Everybody but Mary is eligible. (16 possibilities)
Elect a secretary. Everybody but Mary and the president is eligible. (15 possibilities)
- No
- Is Sam the VP?
Yes (1 possibility) (*)
Elect a president. Everybody but the VP (Sam) is eligible. (16 possibilities)
Elect a secretary. Everybody but the VP and president are eligible. (15 possibilities)
No (15 possibilities)
Elect a president. Everybody but the VP and Sam is eligible. (15 possibilities)
Elect a secretary. Everybody but the VP and president are eligible. (15 possibilities)
- Is Sam the VP?
1\cdot 16 \cdot 15 + 1\cdot 16\cdot 15+15^3=3855 possibilities
(*) This is the case that I overlooked. It may even be that some of you were trying to tell me about it--but I simply didn't see it.
Solution 2: Through the complement, as suggested by Mark
Count all possibilities: 17\cdot 16\cdot 15
Subtract the 'impossible' ones. The implication is 'Sam P -> Mary VP'. The negation of that is 'Sam P and Mary not VP'. Selection procedure:
Elect a president, Sam (1 possibility)
Elect a VP, not Mary or Sam (15 possibilities) (**)
Elect secretary (15 possibilities)
17\cdot 16\cdot 15-1\cdot 15^2=3855 possibilities
(**) Here's another dependendency that might not be easy to spot: If you select the VP first, then you might overlook that Sam cannot be VP. Guess how I know. Combinatorics can be annoying like that.
