How many solutions does the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 28$ have in the non-negative integers?
A particular solution to the equation in the non-negative integers corresponds to the placement of four addition signs in a row of $28$ ones. For instance,
$$1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 + + 1 1 1 1$$
corresponds to the solution $x_1 = 9$, $x_2 = 8$, $x_3 = 7$, $x_4 = 0$, and $x_5 = 4$. Thus, the number of solutions of the equation in the non-negative integers is the number of ways four addition signs can be placed in a row of $28$ ones, which is
$$\binom{28 + 4}{4} = \binom{32}{4}$$
since we must choose which four of the $32$ symbols (four addition signs and $28$ ones) will be addition signs.
In general, the number of solutions of the equation
$$x_1 + x_2 + \cdots + x_k = n$$
in the non-negative integers is equal to the number of ways $k - 1$ addition signs can be inserted into a row of $n$ ones, which is
$$\binom{n + k - 1}{k - 1} = \binom{n + k - 1}{n}$$
since we must choose which $k - 1$ of the $n + k - 1$ symbols ($n$ ones and $k - 1$ addition signs) will be addition signs or, alternatively, which $n$ of the symbols will be ones.
How many solutions does the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 28$ have in the positive integers?
A particular solution of the equation in the positive integers corresponds to the placement of four addition signs in the $27$ spaces between successive ones in a row of $28$ ones. For instance,
$$1 1 1 1 1 1 + 1 1 1 1 1 1 1 + 1 1 1 1 1 1 + 1 1 1 1 1 + 1 1 1 1$$
corresponds to the solution $x_1 = 6$, $x_2 = 7$, $x_3 = 6$, $x_4 = 5$, and $x_5 = 4$. Thus, the number of solutions of the equation in the positive integers is the number of ways four addition signs can be placed in the $27$ gaps between successive ones in a row of $28$ ones, which is
$$\binom{27}{4}$$
In general, the number of solutions of the equation
$$x_1 + x_2 + \cdots + x_k = n$$
in the positive integers is the number of ways $k - 1$ addition signs can be placed in the $n - 1$ gaps between successive ones in a row of $n$ ones, which is
$$\binom{n - 1}{k - 1}$$
How many solutions does the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 28$ have in the integers subject to the restraints $x_k \geq k$, $1 \leq k \leq 5$.
Hint: Let $y_k = x_k - k$, $1 \leq k \leq 5$. Substitute $y_k + k$ for $x_k$, $1 \leq k \leq 5$, then solve the resulting equation in the non-negative integers.
Best Answer
Let us apply the approach suggested in comment.
Let $x_2=x_1+a, x_3=x_2+b$ so that the original equality can be written as: $$ 3x_1+2a+b+x_4+x_5=12 $$ where $a$ and $b$ are positive and the other are non-negative integer numbers.
The number of integer solutions to the equation is: $$ [x^{12}]\frac1{1-x^3}\frac{x^2}{1-x^2}\frac{x}{1-x}\frac{1}{1-x}\frac{1}{1-x}= [x^{12}]\frac{x^3}{(1-x^3)(1-x^2)(1-x)^3}=189. $$ where $[x^n]$ means the coefficient at $x^n$ in the series expansion of the following expression.