Number of ways to write a sum of $3$ numbers that sum to $10$, such that none can be empty? $ \geqslant 1$

combinatorics

How many ways are there of writing the number 10 as
a sum of 3 numbers (none of which may be empty)?

where we can write as this equation $x_1+x_2+x_3=10$, such that $x$'s have to be at least greater than $1$. so is this the same as $x_1+x_2+x_3=7$ where $\binom{3+7-1}{7}$ ?

Best Answer

1st Problem

The number of integer solutions of

$ n_1 + n_2 + \cdots + n_n = m $ (*)

is equivalent to the problem of establishing the number of ways to allocate m balls in n boxes.

Solution

$ C_{n,m}^r = {n + m -1 \choose m}={n + m -1 \choose n-1}$

remembering that $ C_{n,k}^r = {n \choose k}={n \choose n-k}$

2nd Problem

What is the number of integer solutions of (*) so that

$\left\{ \begin{array}{l} n_1 + n_2 + \cdots + n_n = m \\ n_i \geq r_i & i=1,…,n \\ r_i \in Z \\ \end{array} \right. $

Solution

Let $ z_i=n_i-r_i \geq 0$, equation (*) cab be writte as

$ z_1 + z_2 + \cdots + z_n = m – r $ with $z_i \geq 0$ and $r=\sum_{i=1}^n r_i $

Solution

$ C_{n,m}^r = {n + m – r - 1 \choose m-r}={n + m -r - 1 \choose n-1}$

In your case, we have $n=3$, $m=10$, $r_i = 1$ for all $i=1, \cdots, n$ and $r=3$. The answer is:

$ C_{n,m}^r = {3 + 10 – 3 - 1 \choose 10-3}={9 \choose 2}$

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