Finding Number Of Solutions Of Equations - Practice Questions & MCQ

To find the solutions of $\mathrm{a}+\mathrm{b}+\mathrm{c}=6$ such that $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are non-negative integers
Since zero is included, we have $a \geq 0, b \geq 0, c \geq 0$
Several solutions of $a+b+c=6$ are equivalent to the number of ways of distributing 6 identical things into 3 different people, which can be achieved by arranging 6 objects and $3-1=2$ partitions in a row. The number of objects before the first partition is given to the first person, the number of objects in between the two partitions is given to the second person and several objects after the second partition are given to the third person.

Number of ways of doing this arrangement is $\frac{8!}{6!.2!}$, which can also be written as ${ }^{6+3-1} C_{3-1}$
Hence the total number of solutions $={ }^{6+3-1} \mathrm{C}_{3-1}=28$
Generalized formula
For whole number solutions of $a_1+a_2+a_3+\ldots. .+a_r=n$, we have the formula ${ }^{n+r-1} C_{r-1}$.

Generalized formula

The natural number solutions of $a_1+a_2+\ldots. .+a_r=n$ are ${ }^{n+r-1} C_{r-1}$
Q : Find the Natural number of solutions of $\mathrm{a}+\mathrm{b}+\mathrm{c}=6$ such that $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are natural numbers
Solution:
Since zero is included, we have $a \geq 1, b \geq 1, c \geq 1$
Let us define new variables

$
\begin{aligned}
& a^{\prime}+1=a \\
& b^{\prime}+1=b \\
& c^{\prime}+1=c
\end{aligned}
$
Now if $\mathrm{a}^{\prime}, \mathrm{b}^{\prime}, \mathrm{c}^{\prime}$ are whole numbers then $\mathrm{a}, \mathrm{b}, \mathrm{c}$ will be natural numbers.
So the equation becomes $a^{\prime}+1+b^{\prime}+1+c^{\prime}+1=6$

$
a^{\prime}+b^{\prime}+c^{\prime}=3
$
So whole number solutions of this equation equal natural number solutions of $a+b+c=6$
Hence the total number of solutions $={ }^{3+3-1} \mathrm{C}_{3-1}=10$