A high school chess club has 12 members. In how many different ways can they choose a president and vice president?
A. 144
B. 66
C. 23
D. 132

Question

A high school chess club has 12 members. In how many different ways can they choose a president and vice president?
  		A. 	144
  		B. 	66
  		C. 	23
  		D. 	132

solomonzelman · Answer

In this case it would be permutations, because you when you choose person D and person E they can be rearranged in 2 different ways, D is ores and E is vice or the other way.

$\Large\color{blue}{   ~_{12}~P~~_2~=~ \frac{12!}{(12-2)!} }$

anonymous · Answer

@SolomonZelman can i just have the answer please

solomonzelman · Answer

$\Large\color{black}{    \frac{12!}{(12-2)!} =\frac{12!}{10!}=\frac{10!\times 11 \times 12}{10!}  =11 \times 12=?}$

solomonzelman · Answer

just do the last step:)