The number of integer solutions to

x1 + x2 + x3 = 6

with x1,x2,x3 >= 1 is equal to the number of integer solutions to

x1 + x2 + x3 = 3

with x1,x2,x3 >=0.

It is like distributing three similar balls into three different buckets, solutions are:

b b b | | ( x1=3, x2=0, x3=0 )

b b | b | ( x1=2, x2=1, x3=0 ).

A pattern appears. We are in fact looking at the number of different arrangements of three b's and 2 |'s, which is ( 3 + 2 )! / 3! 2! = C(5,2) = 10.

Let's list the solutions anyway.

3 - 0 - 0

2 - 1 - 0

2 - 0 - 1

1 - 2 - 0

1 - 1 - 1

1 - 0 - 2

0 - 3 - 0

0 - 2 - 1

0 - 1 - 2

0 - 0 - 3

or translated to the original question:

4 - 1 - 1

3 - 2 - 1

3 - 1 - 2

2 - 3 - 1

2 - 2 - 2

2 - 1 - 3

1 - 4 - 1

1 - 3 - 2

1 - 2 - 3

1 - 1 - 4.

They are in fact all the ordered 3-partitions of 6.

So we can interpret C(n,k) as the ordered (k+1)-partitions of (n+1).

1-2017 More on the randomness of randomness.

2 months ago

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