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:


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*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.

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*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.
