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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?

Z. Andrea Durio
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1 Answers1

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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.

copper.hat
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  • But won't the choice of the center and the point guard affect the choice of the forwards....as the set you get to choose the forwards changes based on it ....(in b) – Saket Gurjar May 13 '20 at 22:27
  • No. The question ask how many ways... after the centre and the point guard are selected. – copper.hat May 13 '20 at 23:27
  • Wait I'm confused. Why did you choose a combination for b and a permutation for c? The teacher labeled this a permutations problem but should it have been a combination problem? – Z. Andrea Durio May 14 '20 at 14:05
  • Unless you distinguish the two forwards it is a combination. – copper.hat May 14 '20 at 14:22
  • Okay, now I understand. b is the only one that is using the same position twice, which means there is not distinction. Thank you! – Z. Andrea Durio May 14 '20 at 15:25