Question

The number of integral solution of x + y + z = 0 with $x\ge -5,y\ge -5,z\ge -5$ is
a. 134
b. 136
c. 138
d. 140

Answer 1

Hint: In order to solve this question, we will try to reform the equation x + y + z = 0 as a + b + c = n, for a, b, c > 0 and n > 0. And then, we will apply the formula of combination which is used to find the number of ways of choosing values of r terms whose sum is n, that is, for ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......+{{x}_{r}}=n$, the number of ways of choosing ${{x}_{1}},{{x}_{2}},{{x}_{3}}......{{x}_{r}}$ is given by $^{n+r-1}{{C}_{r-1}}$.

Complete step-by-step answer:
In this question, we have been asked to find the number of integral solutions of x + y + z = 0 with $x \ge -5, y \ge -5, z \ge -5$. To solve this question, we will try to reform the equation x + y + z = 0 as a + b + c = n.
We have been given that $x \ge -5, y \ge -5, z \ge -5$. So, we can write them as $x+5 \ge 0, y+5 \ge 0, z+5 \ge 0$.
Now, let us consider x + 5 = a, y + 5 = b and z + 5 = c. Therefore, we can say that $a\ge 0, b\ge 0$ and $c\ge 0$.
So, we can write x = a – 5, y = b – 5 and z = c – 5. Hence, we can write the equation, x + y + z = 0 as a – 5 + b – 5 + c – 5 = 0. And it can be further written as,
a + b + c – 15 = 0
a + b + c = 15
Now, we know that for ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......+{{x}_{r}}=n$, the number of ways of choosing the values of ${{x}_{1}},{{x}_{2}},{{x}_{3}}......{{x}_{r}}$ is calculated by using the formula, $^{n+r-1}{{C}_{r-1}}$. Hence, we can calculate the number of ways of choosing the values for a, b and c for a + b + c = 15 by substituting n = 15 and r = 3 in the formula $^{n+r-1}{{C}_{r-1}}$. So, we will get,
$\begin{align}
  & ^{15+3-1}{{C}_{3-1}} \\
 & ^{17}{{C}_{2}} \\
\end{align}$
Now, we know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. So, for n = 15 and r = 2, we can write,
$^{17}{{C}_{2}}=\dfrac{17!}{2!\left( 17-2 \right)!}$
Now, we will simplify it. So we will get,
$\begin{align}
  & \Rightarrow \dfrac{17!}{2!15!} \\
 & =\dfrac{17\times 16\times 15!}{\left( 2\times 1 \right)15!} \\
 & =\dfrac{17\times 16}{2} \\
 & =17\times 8 \\
 & =136 \\
\end{align}$
Hence, the number of ways of choosing integral values for x + y + z = 0 with $x,y,z\ge -5$ is 136.
Therefore, option (b) is the correct answer.

Note: The possible mistake one can make while solving this question is by applying the formula $^{n+r-1}{{C}_{r-1}}$ for x + y + z = 0, which is wrong because x, y, z are not greater than or equal to 0. This formula is applicable only for $x,y,z\ge 0$ and n > 0, which is 0 in this question.