I am a middle school math teacher helping my niece with permutations. Her teacher gave her a problem and later gave her the solution to the problem. I do not agree with the way her teacher solved it and I would like guidance.
Here is the problem:
Twelve people try out for a basketball team. In how many ways could they select:
a. a center and a point guard?
b. a forward and another forward after the cent and the point guard have been selected?
c. a second guard and an alternative after the center, point guard, and two forwards have been selected?
Here's how her teacher solved it: a. P(12,2)=132 b. [P(12,1)*P(12,1)]-P(12,2) --> 144-132=12 c. [P(12,1)*P(12,1)]-5 --> 144-5=139
Here's How I solved it: a.P(12,2)=132 b.n-->12-2=10. P(10,2)=90 c.n-->12-4=8. P(8,2)=56
Can you tell me which is correct and why and what the error was with the incorrectly solved problem?
For a. You are going to label one a centre and another a point guard, there are $12 \cdot 11 = 132$ ways of doing this.
For b. you have $10$ remaining and you wish label two of the $10$ as forwards, there are $\binom{10}{2} = {10 \cdot 9 \over 1 \cdot 2} = 45$ ways of doing this. Note that there is no distinction between the two forwards, hence a combination, if I select Players $5,9$ that is the same as selecting Players $9,5$.
(Alternatively could view this as labeling two remaining players in which case there are $10 \cdot 9$ ways of doing this, and noticing that each couple will be double counted so I need to divide by $2$).
For c. you have $8$ remaining and you want to label one as a centre and one as an alternative, there are $8 \cdot 7 = 56$ ways of doing this.
I can't even guess what the teacher was trying to do with b & c.