Question 867481
There are two women: Brenda and Dorothy (B, D)



There are three men: Allen, Chad, and Eric (A,C,E)



There are two ways to choose the president. This is because "the president must be a woman".



So we could start off with Brenda (B) to have B __  __ 



The blank spaces represent slots for the secretary and treasurer.



There are 3 men to fill the second slot, then 3-1=2 men to fill the next slot. So 3*2 = 6 permutations possible. These permutations are:



BAC
BCA
BAE
BEA
BCE
BEC



Note: order matters and BAC is different from BCA (since Allen is secretary and Chad is treasurer in BAC; but Chad is secretary and Allen is treasurer in BCA)



If we picked Dorothy for president, then we get



DAC
DCA
DAE
DEA
DCE
DEC



There are 6 ways in the first group (with Brenda as president) and there are 6 ways in the second group (with Dorothy as president). 



So there are 6+6 = <font color="red">12</font> possible ways to have a president, a secretary, and a treasurer, if the president must be a woman and the other two must be men (given the candidates Allen,Brenda,Chad,Dorothy,Eric).