How many ways are there to put 5 balls in 3 boxes if the balls are not distinguishable but the boxes are?
Since the balls are indistinguishable, we must only count the number of balls in the different boxes.

There are $3$ ways to arrange the balls as $(5,0,0)$ (specifically, box 1 can have 5, box 2 can have 5, box 3 can have 5).

There are $3! = 6$ to arrange $(4,1,0)$ and $3! = 6$ ways to arrange $(3,2,0)$; in each case, we must choose one of the 3 boxes to have the largest number of balls, and also one of the remaining two boxes to be left empty.

However, there are only $3$ ways to arrange $(3,1,1)$, and $3$ ways to arrange $(2,2,1)$; in each case, we must choose one box to have the `different' number of balls (3 in the $(3,1,1)$ case and 1 in the $(2,2,1)$ case).

This gives a total of $3 + 6 + 6 + 3 + 3 = \boxed{21}$ arrangements.
21
Level 4
Counting & Probability
Problem: How many ways are there to put 5 balls in 3 boxes if the balls are not distinguishable but the boxes are?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem of distributing 5 indistinguishable balls into 3 distinguishable boxes, we can use the combinatorial method known as "stars and bars." This method allows us to find the number of ways to place dividers (bars) among the balls (stars) to separate them into groups corresponding to the boxes.

First, we represent the 5 balls as stars: *****. We need to place 2 bars to create 3 sections (for the 3 boxes). The total number of symbols (stars plus bars) is 5 stars + 2 bars = 7 symbols.

Next, we need to determine how many ways we can arrange these 7 symbols, where 5 are identical stars and 2 are identical bars. The formula for the number of arrangements of n items where there are groups of identical items is given by the multinomial coefficient:

Number of arrangements = n! / (k1! * k2! * ... * kr!)

In our case, n = 7 (total symbols), k1 = 5 (stars), and k2 = 2 (bars). Thus, we calculate:

Number of arrangements = 7! / (5! * 2!)

Calculating the factorials, we have:

7! = 5040
5! = 120
2! = 2

Now substituting these values into the formula:

Number of arrangements = 5040 / (120 * 2) = 5040 / 240 = 21

Therefore, the total number of ways to distribute 5 indistinguishable balls into 3 distinguishable boxes is 21.<|END_OF_LATENT|> The final answer is $21$.