+52
In how many ways can 4 balls be placed in 3 boxes if the balls are distinguishable, and the boxes are indistinguishable?
+1908 We can use casework to solve this problem.
First, le'ts consider cases with \(4-0-0\), where all 4 balls are in one box.
Since each box is indistinguiashable, there is only 1 way to complete this.
Second, let's consider cases with \(3-1-0\)
We can calculate this by using \(4 \choose 1 \), so there are 4 ways.
Third, let's consider cases with \(2-2-0\)
There are \(4 \choose 2\) ways to complete this, so there are 6 cases.
Lastly, we have \(2-1-1\)
We have \(4 \choose 2 \) ways, to do this, so 6 additional cases.
Now, we add all of these up. We get \(1 + 4 + 6 + 6 = 17\)
So our final answer is 17.
Thanks! :)
+52 Sorry, that is not correct. Thanks for trying.
The correct solution is this:
For the 4, 0, 0 distribution we have only 1 way.
For the 3, 1, 0 distribution we have 4 ways.
The 2, 2, 0 distribtution is where NotThatSmart makes a mistake. It would appear to be \(4 \choose 2\) = 6 ways, but there is double counting. For example, choosing balls 1 and 3 leads to the same split as choosing balls 2 and 4. So, for the 2, 2, 0 distribution there are only 3 ways, making the total amount 14 ways.
+1908 I see what you mean now.
Instead of it being \(4\choose 2\), we have overcounting to consider.
Now, we only have 3 cases since there is no distinguishbility for the boxes.
Let's name the balls ball 1, 2, 3, and 4.
We have
1/2 and 3/4
1/3 and 2/4
1/4 and 2/3
We already accounted for anyother case.
Thank you for the correction, MEMEGOD!
Thanks! :)
