Counting Possibilities in Table Tennis and 3-Digit Numbers

[L²Cc]

Homework Statement

In how many different ways can the top two positions be filled in a table tennis competition of 7 teams?
How many 3-digit numbers can be formed using the digits 2, 3, 4, 5, and 6:
a) as often as desired? b) once only?

Homework Equations

The Attempt at a Solution

I can find the solution using the long way, but I'm wondering if there is a more logical and mathematical way to approach these equations?

 

[rootX]

you know permutations and combinations?

 

[L²Cc]

I got the first equation, and the b) section of equation 2, but how about a)? I'm confused by what they are asking.

 

[L²Cc]

yes...that's what I used to solve for equation 1, and b), but a's wording is slightly confusing :p

 

[nrqed]

yes...that's what I used to solve for equation 1, and b), but a's wording is slightly confusing :p

You may use the same digit twice or even three times (like 665 or 666). That's all they mean.
So how many choices do you have for the first digit? how many choices for the second? How many for the third? Multiply those three numbers and you are done.

 

[L²Cc]

Oh Ok. So just to clarify things, each number could be used five times...therefore 5^3...and when each value is used once, 5x4x3...which suggests that in the first box, any value of the series could be used, in the second box, what's left over, etc...is my explanation correct? If you have a better one, feel free to share it ;p.

 

[HallsofIvy]

Oh Ok. So just to clarify things, each number could be used five times...therefore 5^3...and when each value is used once, 5x4x3...which suggests that in the first box, any value of the series could be used, in the second box, what's left over, etc...is my explanation correct? If you have a better one, feel free to share it ;p.

Please be careful of your wording: each digit (better than "number" which is confusing) can be used three times. What you really mean to say, I think, is that for each digit in the 3 digit number, you can choose from any of the 5 given digits. Yes, you are correct that then there are 5³= 125 possible numbers. If you are only allowed to use each of the given digits once, then there are 5 choices for the first digit, 4 choices for the second digit, 3 choices for the third digit and so 5(4)(3)= 60 possible numbers.