How many ways are there to place 7 distinct balls into 3 distinct boxes?

Question

How many ways are there to place $7$ distinct balls into $3$ distinct boxes?

is the question I'm confused about.

The solution shows that the correct answer is $3^7$. I'm just confused why this is.

My thinking is that if there are 3 boxes, and 7 possible balls for each box:

number of choices:  7 6 5
individual boxes:  _ _ _

So $7*6*5$ total possibilities...

But clearly, the logic in this problem is the following:

Number of choices: 3 3 3 3 3 3 3
Individual balls:   _ _ _ _ _ _ _

Why is the 1st solution incorrect?

Answer 1

We are given the task of placing 7 balls into 3 jars. Step 1: Place 1st ball, 3 ways to do that. Step 2: Place 2nd ball, 3 ways to do that....Step 7: place last(seventh) ball, 3 ways to do that. By rule of product, we have $3*3*3*3*3*3*3 = 3^7$ ways to accomplish the task. Your method is wrong because assumes we need to put a ball in the first jar. We don't need to put anything in the first jar.

Answer 2

If we go crazy and try to solve it the hard way by considering what each box contains, we could do as follows:

1. Box 1 first chooses a number $0\leq a\leq 7$.
2. Then box 2 gets to choose a number $0\leq b\leq 7-a$.
3. Finally box 3 is forced to choose $c=7-a-b$.
4. Then box 1 chooses $a$ balls in $\binom{7}{a}$ ways.
5. Next box to chooses $b$ of the remaining balls in $\binom{7-a}{b}$ ways.
6. Finally box 3 is forced to accept the remaining $7-a-b$ balls.

Then we could proceed to make a difficult looking sum of cases and numbers of possibilities:

$$ 3^7=\sum_{a=0}^7\sum_{b=0}^{a-7}\binom{7}{a}\binom{7-a}{b} $$

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The simple way, as I see it, is to assign a number to each ball from the set $\{0,1,2\}$. Then a combination of placing the balls corresponds exactly to a seven digit ternary number. Of course there will be $3^7$ of those :-)

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This is so easily generalized to $n$ digit base $k$ numbers corresponding to placing $n$ balls in $k$ boxes, thus making $k^n$ setups.