Consider the equation $x_1+x_2+x_3+x_4 = 35$. How many different solutions does this equation have if all the variables must be positive integers?

Question

Consider the equation $x_1+x_2+x_3+x_4 = 35$. How many different solutions does this equation have if all the variables must be positive integers?

I don't understand this problem.I don't know whether to solve it using bit strings concept or combinations. Someone help!

Answer 1

Let us take each $x_i=a_1+1$

$x_1+x_2+x_3+x_4=35$
$\therefore a_1+a_2+a_3+a_4=31$ ,where each $a_i$ is non-negative.

Now using the stars and bars method the answer is - $\binom{31+4-1}{4-1}\\ =\binom{34}{3}$

Answer 2

For reasons of space, let's consider the following alternative problem:

How many solutions does the equation $x_1 + x_2 + x_3 + x_4 = 15$ have in the positive integers?

A particular solution of this equation corresponds to the placement of $3$ addition signs in the $14$ spaces between successive ones in a row of $15$ ones. $$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$ For instance, placing an addition sign in the fourth, seventh, and twelfth spaces gives $$1 1 1 1 + 1 1 1 + 1 1 1 1 1 + 1 1 1$$ which corresponds to the solution $x_1 = 4$, $x_2 = 3$, $x_3 = 5$, $x_4 = 3$. Hence, the number of such solutions is the number of ways we can select which three of the $14$ spaces between successive ones in a row of $15$ ones will be filled with addition signs, which is $$\binom{14}{3}$$

How many solutions does the equation $x_1 + x_2 + x_3 + \cdots + x_k = n$ have in the positive integers?

Since a particular solution of the equation corresponds to the placement of $k - 1$ addition signs in the $n - 1$ spaces between successive ones in a row of $n$ ones, the number of solutions of the equation $x_1 + x_2 + x_3 + \cdots + x_k = n$ in the positive integers is $$\binom{n - 1}{k - 1}$$