# Write n = a + b + c + ...

#### hxthanh

##### New member
$\boxed 1$ How many ways writing positive integer $n\;$ as the sum of the positive integers different each pairs? (no permutation)

Example: $6=6=1+5=2+4=1+2+3 \quad$ (4 ways)

$\boxed 2$ How many ways writing positive integer $n\;$ as the sum of the positive integers? (no permutation)

Example: \begin{align*} 6&=6\\ & =1+5 = 2+4 = 3+3 \\&= 1+1+4 =1+2+3 =2+2+2\\ &=1+1+1+3 =1+1+2+2 \\& =1+1+1+1+2 \\&=1+1+1+1+1+1 \end{align*}\quad (11 ways)

#### soroban

##### Well-known member
Re: Write n=a+b+c+...

If permutations are allowed, the problem is elementary.

Consider a board $$n$$ inches long.
Mark the board in one-inch intervals.

. . . . $$\underbrace{\square\!\square\!\square\!\square \cdots \square}_{n-1\text{ marks}}$$

At each mark, we have 2 choices: cut or do not cut.

Hence, there are $$2^{n-1}$$ possible decisions.

Therefore, there are $$2^{n-1}$$ possible addition problems.

Example: $$n = 4$$
. . $$\begin{array}{c} 4 \\ 1+3, \;\; 2+2, \;\;3+1 \\ 1+1+2, \;\; 1+2+1, \;\; 2 +1+1 \\ 1+1+1+1+1 \end{array}\quad(2^3 = 8\text{ ways)}$$

However, if permutations are not allowed,
. . the problem become very difficult.

Many years ago, my then-wife was in Special Education.
I did some scribbling and came up with the formula.
I proudly announced 2^4 = 16 problems and listed them.

Then she asked how many if permutations were not allowed.
I spent the rest of the weekend and got nowhere. .On Monday,
I gave the problem to my fellow math professors. .Within