Question

The number of ordered triplets of positive integers which are solutions of the equation $x+y+z=100$ is$?$

• A. $6005$
• B. $4851$
• C. $5081$
• D. None of these

Answer 1

The correct option is B

$4851$

Explanation for the correct option:

Find the required number of ordered triplets.

An equation $x+y+z=100$ is given.

Since the sum $n$ of all variables is $100$ and the number $r$ of variables is $3$.

So, the total number of triplets is given by ${}^{n-1}C_{r-1}$.

$$\begin{gathered} {}^{n-1}C_{r-1}={}^{100-1}C_{3-1} \\ \Rightarrow {}^{n-1}C_{r-1}={}^{99}C_2 \\ \Rightarrow {}^{n-1}C_{r-1}=\frac{99·98}{2} \\ \Rightarrow {}^{n-1}C_{r-1}=4851 \end{gathered}$$

Therefore, The number of ordered triplets of positive integers which are solutions of the equation $x+y+z=100$ is $4851$.

Hence, option $B$ is the correct answer.