Question

How many solutions are there to the equation ${\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{3}}\mathbf{+}{\mathbf{x}}_{\mathbf{4}}\mathbf{=}\mathbf{17}$where${\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{3}}\mathbf{+}{\mathbf{x}}_{\mathbf{4}}$are nonnegative integers?

Short answer

There are 1140 solutions to the equation ${\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=17$ where ${\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4$are nonnegative integers.

Step-by-step solution

Step 1: Definitions

Definition of Permutation (Order is important)

No repetition allowed:${}^{P(n,r)=\frac{n!}{(n-r)!}}$

Repetition allowed:$n^T$

Definition of combination (order is important)

No repetition allowed:$C(n,r)=\left(\begin{array}{l}n\\ r\end{array}\right)=\frac{n!}{r!(n-r)!}$

Repetition allowed:$C(n+r-1,r)=\left(\begin{array}{c}n+r-1\\ r\end{array}\right)=\frac{(n+r-1)!}{r!(n-1)!}$

with$n!=n(n-1)\cdot .....\cdot 2\cdot 1$

Step 2: Solution

${\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=17$

The integer solutions of the equationdata-custom-editor="chemistry" ${\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+.............+{\mathrm{x}}_{\mathrm{n}}=\mathrm{r}$ can be obtained by selecting r objects from a set with n objects that there are${\mathrm{x}}_1$ chosen from the first type,${\mathrm{x}}_2$ chosen from second type and so on.

Thus the number of solutions can then be obtained by using the definition of a combination (since order of the solutions is not important) and repetition is allowed (since more than one x, value can take on the same value)

$\begin{array}{r}\mathrm{n}=4\text{and}\mathrm{r}=17\\ \mathrm{C}(\mathrm{n}+\mathrm{r}-1,\mathrm{r})=\mathrm{C}(4+17-1,17)\\ \mathrm{C}(20,17)=\frac{20!}{17!(20-17)!}=\frac{20!}{17!3!}=1140\end{array}$

Thus there are 1140 solutions to the equation${\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=17$ where ${\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4$are nonnegative integers