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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. 1...

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

Explanation

Solution

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 x1+x2+x3+......+xr=n{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......+{{x}_{r}}=n, the number of ways of choosing x1,x2,x3......xr{{x}_{1}},{{x}_{2}},{{x}_{3}}......{{x}_{r}} is given by n+r−1Cr−1^{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≥−5,y≥−5,z≥−5x \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≥−5,y≥−5,z≥−5x \ge -5, y \ge -5, z \ge -5. So, we can write them as x+5≥0,y+5≥0,z+5≥0x+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≥0,b≥0a\ge 0, b\ge 0 and c≥0c\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 x1+x2+x3+......+xr=n{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......+{{x}_{r}}=n, the number of ways of choosing the values of x1,x2,x3......xr{{x}_{1}},{{x}_{2}},{{x}_{3}}......{{x}_{r}} is calculated by using the formula, n+r−1Cr−1^{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−1Cr−1^{n+r-1}{{C}_{r-1}}. So, we will get,
15+3−1C3−1 17C2 \begin{aligned} & ^{15+3-1}{{C}_{3-1}} \\\ & ^{17}{{C}_{2}} \\\ \end{aligned}
Now, we know that nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}. So, for n = 15 and r = 2, we can write,
17C2=17!2!(17−2)!^{17}{{C}_{2}}=\dfrac{17!}{2!\left( 17-2 \right)!}
Now, we will simplify it. So we will get,
⇒17!2!15! =17×16×15!(2×1)15! =17×162 =17×8 =136 \begin{aligned} & \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{aligned}
Hence, the number of ways of choosing integral values for x + y + z = 0 with x,y,z≥−5x,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−1Cr−1^{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≥0x,y,z\ge 0 and n > 0, which is 0 in this question.