# Thread: Permutation and organized counting

1. ## Permutation and organized counting

Hello, I would like some help in solving these problems. Please explain how you arrived at your answer if you don't mind.

1. Ten finalists are competing in a race at the Canada games:
a) In how many different orders can the competitors finish the race?
b) How many ways could the gold, silver, and bronze medal be awarded?
c) One of the finalists is a friend from your town. How many of the possible finishes would include them winning a medal?
d) How many possible finishes would leave your friend out of the medal standings?
e) Suppose one of the competitors is injured and cannot finish the race. How does that affect your previous answers?
f) How would the competitor's injury affect your friend's chances of winning a medal? Explain your reasoning. What assumptions have you made?

2. Originally Posted by Tarwamin
Hello, I would like some help in solving these problems. Please explain how you arrived at your answer if you don't mind.

1. Ten finalists are competing in a race at the Canada games:
a) In how many different orders can the competitors finish the race?
b) How many ways could the gold, silver, and bronze medal be awarded?
c) One of the finalists is a friend from your town. How many of the possible finishes would include them winning a medal?
d) How many possible finishes would leave your friend out of the medal standings?
Do you have any ideas on how to solve these?
This is not a homework service.

3. I understand that this is not a homework service. I have tried the questions, but I am not sure if they're right. I just wanted to confirm by comparing answers.

21a) 10! = 10x9x8x7x6x5....x2x1= 3628800
b) 10x9x8= 720 since there are only 3 positions to fill
c) 9x8x3=216
d) 9x8x7 =504
e) I'm not sure
f) I'm not sure

I'm also having problems explaining HOW I arrived to these answers. I'm just doing it because that's how I was taught to. I don't know how to 'justify' them.

4. Originally Posted by Tarwamin
a) 10! = 10x9x8x7x6x5....x2x1= 3628800
b) 10x9x8= 720 since there are only 3 positions to fill
c) 9x8x3=216
d) 9x8x7 =504
e) I'm not sure
f) I'm not sure

(c) and (d) are more complicated. To prove that your given answers are incorrect, they do not sum to 10! = 3628800. But you must have some reasoning for them.

5. Originally Posted by Tarwamin
I understand that this is not a homework service. I have tried the questions, but I am not sure if they're right. I just wanted to confirm by comparing answers.

21a) 10! = 10x9x8x7x6x5....x2x1= 3628800
b) 10x9x8= 720 since there are only 3 positions to fill
c) 9x8x3=216
d) 9x8x7 =504
e) I'm not sure
f) I'm not sure
You are correct on parts A) & B).
Part C): $\displaystyle _9C_2=\frac{9!}{2!\cdot 7!}$

Part D) $\displaystyle _9C_3$

You posted neither parts E) nor F).

6. Originally Posted by Plato
You are correct on parts A) & B).
Part C): $\displaystyle _9C_2=\frac{9!}{2!\cdot 7!}$

Part D) $\displaystyle _9C_3$

You posted neither parts E) nor F).
Are you sure about this? If your friend wins a medal, then either he placed 1st, 2nd, or 3rd. In each of those cases, the other people can place in 9! different ways.

C) 3 x 9!

D) 7 x 9!

7. Originally Posted by icemanfan
Are you sure about this? If your friend wins a medal, then either he placed 1st, 2nd, or 3rd. In each of those cases, the other people can place in 9! different ways.
C) 3 x 9!
D) 7 x 9!
What in the world makes you think that order has anything to do with this question?
The question is about the ‘friend’ being among the top three metal winners.
That means that there are two of nine with the ‘friend’.
That is a subset count: combinations.