In how many different ways can a dozen identical red chips be placed into three separate piles if each pile must end up with at least one chip?

Question

 In how many different ways can a dozen identical red chips be placed into three separate piles if each pile must end up with at least one chip?  

Answer 1

Hi Christopher,

 

Lets imagine the number of chips in each pile to be expressed as a sequence of three numbers, the first representing the number of chips in pile 1, the second number representing the number of chips in pile 2, and the third representing the number of chips in pile 3. For example, if I put 1 chip in pile 1, 1 chip in pile 2, and 10 chips in pile 3, I would write

 

1  1  10

 

Let's first write every combination where the number of chips in pile 1 is 1:

 

1  1  10

1  2  9

1  3  8

1  4  7

1  5  6

1  6  5

1  7  4

1  8  3

1  9  2

1  10  1

 

This makes 10 possibilities.

 

Let's now write every combination where the number of chips in pile 1 is 2:

 

2  1  9

2  2  8

2  3  7

2  4  6

2  5  5

2  6  4

2  7  3

2  8  2

2  9  1

 

This makes a total of 9 possibilities

 

Let's now write every combination where the number of chips in pile 1 is 3:

 

3  1  8

3  2  7

3  3  6

3  4  5

3  5  4

3  6  3

3  7  2

3  8  1

 

This makes a total of 8 possibilities.

 

We can continue in this pattern for the number of chips in pile 1 to be 4, 5, 6, and so on until we get to 10, and in this case there is only 1 combination:

 

10  1  1

 

So, the total number of combinations of chips in each pile where each pile has a minimum of 1 chip is:

10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55

 

You could also use the Gauss formula for adding the first n numbers starting with 1: n(n + 1)/2 = (10)(11)/2 = 55.