# Math Help - choosing positions

1. ## choosing positions

Ms. G has 8 boys and 6 girls in her class. The class is going to elect a president, vice president, secretary, and treasurer. Tom and Ruth, members of the class are no longer speaking and will not work together. If all members of the class are willing to be considered for any of the offices and any slate of officers with tom and ruth is an unworkable slate, how many workable slates of officers could be considered?

2. Originally Posted by ihavvaquestion
Ms. G has 8 boys and 6 girls in her class. The class is going to elect a president, vice president, secretary, and treasurer. Tom and Ruth, members of the class are no longer speaking and will not work together. If all members of the class are willing to be considered for any of the offices and any slate of officers with tom and ruth is an unworkable slate, how many workable slates of officers could be considered?
The way I see this is first take the number of permutations total which is 14P4 or $\left(14\right)_{4}$, then subtract the permutations of Tom and Ruth which is 4P2 or $\left(4\right)_{2}$.

But since I'm just learning this myself, let's see if the experts agree.

3. Originally Posted by oldguynewstudent
The way I see this is first take the number of permutations total which is 14P4 or $\left(14\right)_{4}$, then subtract the permutations of Tom and Ruth which is 4P2 or $\left(4\right)_{2}$.
The basic concept is correct but not the count.
There are $\binom{12}{2}(4!)$ ways for Tom and Ruth to serve together.

4. Hello, ihavvaquestion!

Ms. G has 8 boys and 6 girls in her class.
The class is going to elect a President, Vice President, Secretary, and Treasurer.
Tom and Ruth, members of the class are no longer speaking and will not work together.

If all members of the class are willing to be considered for any of the offices
and any slate of officers with Tom and Ruth is an unworkable slate,
how many workable slates of officers could be considered?

There are: . $_{14}P_4 \:=\:14\cdot13\cdot12\cdot11 \:=\:24,024$ possible outcomes.

Suppose Tom and Ruth are elected.

Tom can have any of the 4 positions.
Ruth can have any of the other 3 positions.

The other two positions are filled by two of the other 12 students.
. . There are: . $_{12}P_2 \:=\;132$ ways.

Hence, there are: . $4\cdot3\cdot132 \:=\:1584$ unworkable slates.

Therefore, there are: . $24,024 - 1584 \;=\;22,440$ workable slates.

5. why are there 14P4 possible outcomes and not 14C4