How many ways are there to put 5 balls in 3 boxes if the balls are distinguishable but the boxes are not?
Since the boxes are indistinguishable, there are 5 different cases for arrangements of the number of balls in each box: $(5,0,0)$, $(4,1,0)$, $(3,2,0)$, $(3,1,1)$, or $(2,2,1)$.

$(5,0,0)$: There is only $1$ way to put all 5 balls in one box.

$(4,1,0)$: There are $\binom{5}{4} = 5$ choices for the 4 balls in one of the boxes.

$(3,2,0)$: There are $\binom{5}{3} = 10$ choices for the 3 balls in one of the boxes.

$(3,1,1)$: There are $\binom{5}{3} = 10$ choices for the 3 balls in one of the boxes, and we simply split the last two among the other indistinguishable boxes.

$(2,2,1)$:  There are $\binom{5}{2} = 10$ options for one of the boxes with two balls, then $\binom{3}{2} = 3$ options for the second box with two balls, and one option remaining for the third.  However since the boxes with two balls are indistinguishable, we are counting each pair of balls twice, and must divide by two.  So there are $\dfrac{10 \times 3}{2} = 15$ arrangements of balls as $(2,2,1)$.

Thus the total number of arrangements for 3 indistinguishable boxes and 5 distinguishable balls is $1 + 5 + 10 + 10 + 15 = \boxed{41}$.

$\textbf{Alternate solution:}$ There are $3^5 = 243$ arrangements to put 5 distinguishable balls in 3 distinguishable boxes.  Among these 243 arrangements, there is one case in our problem that is counted three times: if all 5 balls are placed in one box and the other two boxes both contain nothing.  This leaves 240 other arrangements.

For every other case, the contents of each box is different, and so these cases are each counted $3! = 6$ times.  Therefore there must be 40 of these cases, and we have $\boxed{41}$ cases total.
41
Level 5
Counting & Probability
Problem: How many ways are there to put 5 balls in 3 boxes if the balls are distinguishable but the boxes are not?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we need to consider the distribution of distinguishable balls into indistinguishable boxes. The first step is to recognize that the arrangement of balls can be represented by partitions of the number 5, where each partition corresponds to a different way of grouping the balls into boxes.

Next, we identify the possible partitions of the number 5. The partitions are as follows:
1. 5 (all balls in one box)
2. 4 + 1 (four balls in one box, one in another)
3. 3 + 2 (three balls in one box, two in another)
4. 3 + 1 + 1 (three balls in one box, one in each of the other two)
5. 2 + 2 + 1 (two balls in two boxes, one in another)

For each partition, we calculate the number of ways to assign the distinguishable balls to the boxes:

1. For the partition 5: There is 1 way to choose all 5 balls to go into one box.
2. For the partition 4 + 1: We choose 1 ball to be alone in a box, which can be done in C(5, 1) = 5 ways.
3. For the partition 3 + 2: We choose 2 balls to go into one box, which can be done in C(5, 2) = 10 ways.
4. For the partition 3 + 1 + 1: We choose 3 balls to go into one box, and the remaining 2 balls each go into their own boxes. The number of ways to choose 3 balls is C(5, 3) = 10 ways.
5. For the partition 2 + 2 + 1: We choose 1 ball to be alone in a box, and the remaining 4 balls are split into two groups of 2. The number of ways to choose 1 ball is C(5, 1) = 5, and the number of ways to split the remaining 4 balls into two groups of 2 is given by (C(4, 2) / 2) = 3 (we divide by 2 to account for the indistinguishability of the two groups).

Now, we sum the number of ways for each partition:
1. 1 way for partition 5
2. 5 ways for partition 4 + 1
3. 10 ways for partition 3 + 2
4. 10 ways for partition 3 + 1 + 1
5. 15 ways for partition 2 + 2 + 1

Calculating the total:
1 + 5 + 10 + 10 + 15 = 41.

Thus, the final answer is 41.<|END_OF_LATENT|> The final answer is $41$.