# Thread: Game Show Possibilities Problem...

1. ## Game Show Possibilities Problem...

Hi Everyone,

Would the following problem be a permutation, or a combination?

"For a gameshow, there are 9 spots to select on a board. You get to choose 3 of them in order to try to find the grand prize. How many different ways are there to select your spots?"

I think it's a permutation, because order matters. In that case, can I still solve it using 9C3 (9 choose 3)? Or do I need to do something else to account for the position of the answers?

Also, this one:

"If a Congressional committee is to be made of six people selected from a group of eight Democrats and five Republicans, how many ways are there to select exactly one Republican?"

The part that throws me off here is the "exactly one" republican. It wouldn't be as simple as a 13C6 calculation, but would that be part of it?

Thank you!
-matt

2. ## Re: Game Show Possibilities Problem...

Originally Posted by BackstreetZAFU
"For a gameshow, there are 9 spots to select on a board. You get to choose 3 of them in order to try to find the grand prize. How many different ways are there to select your spots?"

I think it's a permutation, because order matters. In that case, can I still solve it using 9C3 (9 choose 3)? Or do I need to do something else to account for the position of the answers?
There is no way to answer this unless we know the exact way the game is played. What are the rules?

Originally Posted by BackstreetZAFU
"If a Congressional committee is to be made of six people selected from a group of eight Democrats and five Republicans, how many ways are there to select exactly one Republican?"
This is multiplicative: $\dbinom{5}{1}\cdot \dbinom{8}{5}$

3. ## Re: Game Show Possibilities Problem...

[QUOTE=Plato;946524]There is no way to answer this unless we know the exact way the game is played. What are the rules?

I'm assuming the problem means that you choose three spots (out of 9), and then reset. So, I could choose spots 1, 2, and 3 (in that order), and then reset. Then I could choose spots 1, 3, and 2 (in that order), and reset. And so on. Does that make more sense?

Regarding the second problem, is that the same as 5C1 x 8C5?

-matt

4. ## Re: Game Show Possibilities Problem...

Also, regarding the Congressional problem, if it read:

Using the situation from above, how many ways are there to select exactly three Republicans?

Would it be?

5 Choose 3 x 8 Choose 3
5C3 x 8C3 =
10 x 56 = 560

Thank you again!
-matt

5. ## Re: Game Show Possibilities Problem...

[QUOTE=BackstreetZAFU;946527]
Originally Posted by Plato
I'm assuming the problem means that you choose three spots (out of 9), and then reset. So, I could choose spots 1, 2, and 3 (in that order), and then reset. Then I could choose spots 1, 3, and 2 (in that order), and reset. And so on. Does that make more sense?
NO! Sorry that makes no sense at all.
Let this be a lesson to you for future posts. If you are assigned a question then post the exact, Wording of that question.
If you post your understanding of the question it is possible that any confusion is at least doubled.
So post the exact wording of a question. Post your understanding of that question. Post your efforts to solve that question.

6. ## Re: Game Show Possibilities Problem...

Originally Posted by BackstreetZAFU
Hi Everyone,

Would the following problem be a permutation, or a combination?
It is better not to worry about that at all!

"For a gameshow, there are 9 spots to select on a board. You get to choose 3 of them in order to try to find the grand prize. How many different ways are there to select your spots?"
You have 9 choices for the first spot. After you have made that choice there are 8 spots left so you have 8 choices for the second spot. Then there are 7 spots left for the third choice.. You can do this in 9*8*7= 504 ways. That is the same as $\displaystyle \frac{9*8*7*6*5*4*3*2*1}{6*5*4*3*2*1}= \frac{9!}{6!}= \frac{9!}{(9- 3)!}= P(9, 3)$ but it is better to think about a problem than to memorize formulas.

I think it's a permutation, because order matters. In that case, can I still solve it using 9C3 (9 choose 3)? Or do I need to do something else to account for the position of the answers?

Also, this one:

"If a Congressional committee is to be made of six people selected from a group of eight Democrats and five Republicans, how many ways are there to select exactly one Republican?"

The part that throws me off here is the "exactly one" republican. It wouldn't be as simple as a 13C6 calculation, but would that be part of it?

Thank you!
-matt
Start by choosing the republican first- there are 5 ways to do that. Now choose the first democrat. There are 8 ways to do that. That leaves 7 democrats so there are 7 ways to choose the second democrat. There are then 6, 5, and 4 ways to choose the third, fourth, and fifth democrats. The total number of ways to choose these 6 people, in this order, is 5*8*7*6*5*4= 5*8!. However, the order in which we choose the people are chosen is not relevant so we divide by 6! to remove the 6! different orders:
$\displaystyle \frac{5*8!}{6!}= 5*8*7= 280$.

7. ## Re: Game Show Possibilities Problem...

[QUOTE=Plato;946539]
Originally Posted by BackstreetZAFU
NO! Sorry that makes no sense at all.
Let this be a lesson to you for future posts. If you are assigned a question then post the exact, Wording of that question.
If you post your understanding of the question it is possible that any confusion is at least doubled.
So post the exact wording of a question. Post your understanding of that question. Post your efforts to solve that question.
That WAS the exact wording of the question. Sorry you couldn't figure it out.

8. ## Re: Game Show Possibilities Problem...

Originally Posted by BackstreetZAFU
Regarding the second problem, is that the same as 5C1 x 8C5?
Huh? Where's the first problem???