Without complement solve combinatorics problem For the question

A Math Club consists of $16$ males and $15$ females. How many ways can the club elect a president, vice president, and treasurer if at least one officer must be female?

The way I tried to do this problem is that there are three cases:


*

*i) If one officer is female: $15 \cdot {_{16}P_2}$

*ii) If two officers are female: ${_{15}P_2} \cdot {_{16}P_1}$

*iii) If three officers are female: ${_{15}P_3}$
Then I add the three cases $15 \cdot {_{16}P_2} + {_{15}P_2} \cdot {_{16}P_1} + {_{15}P_3} = 9,690$ but the answer is wrong because when I use the complement I get ${_{31}P_3} - {_{16}P_3} = 23,610$. I want to learn how to get the right answer without using the complement and find out what is wrong in my logic above.
 A: *

*To i): 
There are $3$ such possibilities: 

$1$ female may be in the role of president, vice president, or treasurer.





*

*To ii):
There are $3$ such possibilities: 

$2$ females may be in the roles of president and vice president, or president and treasurer, or vice president and treasurer.





*

*To iii):
It's OK, there is only one such possibility:

So your expressions both in i) and in ii) multiply by $3$, and you will obtain the right answer.
A: There are three basic cases, all officers are female, two officers are female, and one officer is female.  
When all are female the number of possibilities is $15\times14\times13=2730$.
When two are female the number of possibilities is $15\times14\times16\times3=10080$.
When one officer is female the number of possibilities is $15\times16\times15\times3=10800$.
Sum to get $23610$.
A: 
The way I tried to do this problem is that there are three cases:

*

*i) If one officer is female: $15 \cdot {^{16}\mathrm P_2}$

Which officer?  You want to count ways to select one female officer, one position for her, and select and arrange two male officers. $${^{15}\mathrm P_1}\cdot 3\cdot{^{16}\mathrm P_2}$$


*

*ii) If two officers are female: ${^{15}\mathrm P_2} \cdot {^{16}\mathrm P_1}$

Second verse, same as the first; this time selecting a position for the lone male  $${^{15}\mathrm P_2} \cdot {^{16}\mathrm P_1}\cdot 3$$


*

*iii) If three officers are female: ${^{15}\mathrm P_3}$

... and that is okay.

Then I add the three cases $15 \cdot {^{16}\mathrm P_2} + {^{15}\mathrm P_2} \cdot {^{16}\mathrm P_1} + {^{15}\mathrm P_3} = 9,690$ but the answer is wrong because when I use the complement I get ${^{31}\mathrm P_3} - {^{16}\mathrm P_3} = 23,610$. I want to learn how to get the right answer without using the complement and find out what is wrong in my logic above.

And lo, behold that: ${^{15}\mathrm P_1}\cdot 3\cdot{^{16}\mathrm P_2}+{^{15}\mathrm P_2}\cdot{^{16}\mathrm P_1}\cdot 3+{^{15}\mathrm P_3}=23,610$
