Question

Number of positive integral solutions of the equation $x+y+z=10$ are
(A) $^{14}{{C}_{3}}$
(B) $^{9}{{C}_{2}}$
(C) $^{7}{{C}_{3}}$
(D) None of these

Answer 1

For answering this question we have been asked to find the number of positive integral solutions of the equation $x+y+z=10$ . For doing that we will use the statement “The number of positive integral solutions of the equation ${{x}_{1}}+{{x}_{2}}+.......+{{x}_{r}}=n$ is given as $^{n-1}{{C}_{r-1}}$”.

Complete step-by-step solution:
Now considering from the question we have been asked to find the number of positive integral solutions of the equation $x+y+z=10$ .
For doing that we will use the following statement which we have learnt during the basic concepts of combinations given as “The number of positive integral solutions of the equation ${{x}_{1}}+{{x}_{2}}+.......+{{x}_{r}}=n$ is given as $^{n-1}{{C}_{r-1}}$” . Here, positive solutions indicate that r is greater than 0. So, r = 1, 2, 3, ...
By applying this concept here we will have ${{\Rightarrow }^{10-1}}{{C}_{3-1}}{{=}^{9}}{{C}_{2}}$ .
Therefore we can conclude that the number of positive integral solutions of the equation $x+y+z=10$ is $^{9}{{C}_{2}}$ . Therefore the correct option is “B”.

Note: While answering questions of this type we should be sure with our concepts that we are going to apply. This is a very simple question if we are aware of this concept. Similarly we have learnt another statement during learning the basics of permutations and combinations which is given as “The number of non-negative integral solutions of the equation ${{x}_{1}}+{{x}_{2}}+.......+{{x}_{r}}=n$ is given as $^{n+r-1}{{C}_{r-1}}$” . Here, non-negative solutions indicate that they are greater than or equal to zero. So, r = 0, 1, 2, 3, …. We also know that the value of $^{n}{{C}_{r}}$ is given as $\dfrac{n!}{\left( n-r \right)!}$ . Hence the simplified answer will be given as $\dfrac{9!}{6!}=9\times 8\times 7\Rightarrow 504$ .