Understanding Combinations with Replacement in Probability

[hotvette]

1. Homework Statement

If a total of 5 distinct awards are distributed among 30 students where any student can receive more than 1 award, how many possible outcomes are there?

2. Homework Equations
\text{outcomes} = r^n
where r is the number of choices and n is the number of draws.

3. The Attempt at a Solution

I know the answer is 30^5 but I don't see why 30 is the number of choices and 5 is the number of draws. I know in the case of how many numbers can be formed using 8 binary digits is 2^8. In this case it kind of makes sense that there are only two choices (0 or 1) and I perform the operation 8 times, but with the award and student problem is confusing to me. I just don't see how 30 is the number of choices. I can just as easily say 5 is the number of choices; a student can have up to 5 awards. Only the other hand, only 1 student could get 5 awards, so it really isn't the same as the binary number problem.

What is the thought process to properly sort this out? I've also seen explanations in terms of bins and balls but it's tough to figure out which is the bin and which is the ball. There is something conceptually I'm not getting. Can someone explain?

 

[pat8126]

Each time a medal is awarded, there are 30 possible students that receive it. Based on that possible outcome, there are then 30 more possible more outcomes relative to the next medal, etc.

30 possible students get the 1st medal of 5 medals = 30 possible outcomes
30 possible students get the 2nd medal of 5 medals = 30 possible outcomes
30 possible students get the 3rd medal of 5 medals = 30 possible outcomes
30 possible students get the 4th medal of 5 medals = 30 possible outcomes
30 possible students get the 5th medal of 5 medals = 30 possible outcomes

Therefore, it's 30 x 30 x 30 x 30 x 30 possible outcomes.

Sometimes it helps to view it as a branching tree. Assume 3 students (A, B, C) and 3 medals (1, 2, 3).

(A1, B1, C1) is the first set of possibilities
Then, relative to each of those possibilities, (A2, B2 or C2) is the next branch
Then, relative to each of those possibilities, (A3, B3, C3) for the next branch

A1, A2, A3
A1, A2, B3
A1, A2, C3

A1, B2, A3
A1, B2, B3
A1, B2, C3

A1, C2, A3
A1, C2, B3
A1, C2, C3

B1, A2, A3
B1, A2, B3
B1, A2, C3

B1, B2, A3
B1, B2, B3
B1, B2, C3

B1, C2, A3
B1, C2, B3
B1, C2, C3

C1, A2, A3
C1, A2, B3
C1, A2, C3

C1, B2, A3
C1, B2, B3
C1, B2, C3

C1, C2, A3
C1, C2, B3
C1, C2, C3

 

[hotvette]

Geesh, it seems too easy when explained clearly. Thanks!.

 

[pat8126]

Geesh, it seems too easy when explained clearly. Thanks!.

You are welcome. I'm glad it makes sense.