# 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.