How many ways are there to place 7 distinct balls into 3 distinct boxes? 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?
 A: 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.
A: If we go crazy and try to solve it the hard way by considering what each box contains, we could do as follows:


*

*Box 1 first chooses a number $0\leq a\leq 7$.

*Then box 2 gets to choose a number $0\leq b\leq 7-a$.

*Finally box 3 is forced to choose $c=7-a-b$.

*Then box 1 chooses $a$ balls in $\binom{7}{a}$ ways.

*Next box to chooses $b$ of the remaining balls in $\binom{7-a}{b}$ ways.

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

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 :-)

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