Combinatory problem

Try and think of it this way.

Place the first ball - you have n choices

Now when you place the second ball you still have n choices as there is no restriction on the number of balls in each box.

For example if n = 3 balls. Let's place only balls 1 and 2 and we will not place ball 3.

So for the ball1 you have 3 choices and for each of those 3 choices we have 3 choices for ball2. That's why we multiply 3x3 to get 9 different ways for the first 2 balls.

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Place ball1 (3 choices)

box1 box2 box3

1 -- --

-- 1 --

-- -- 1

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Now place ball 2. Here are all 9 ways.

These are the 3 ways from green above - box1 has ball1. Note that ball 2 has 3 spots

box1 box2 box3

1 2 --

1 -- 2

1,2 -- --

These are the 3 ways from orange above - box2 has ball1. Again ball2 has 3 spots.

box1 box2 box3

2 1 --

-- 1 2

-- 1,2 --

These are the 3 ways from yellow above - box3 has ball1. Again ball2 has 3 spots.

box1 box2 box3

-- 2 1

2 -- 1

-- -- 1,2

Now we have to place ball3.

I'm not going to write them all out, so you are going to have to imagine that for each of the 9 lines in the ball2 section (3G,3O,2Y) you will generate 3 different ways of placing ball3 giving us a total of 9x3 = 27 ways.

The green line 1,2,-- will give us (13,2,--) (1,23,--) (1,2,3)

Repeat 8 more times

So we have 3^3 ways of placing the 3 balls into the 3 boxes.

So for each of the n balls we have n choices for each ball ====> n*n*n*n*...*n (n times) = n^n