Question

The number of solutions of the equation $x+y+z=10$ in positive integers $x,y,z$ is equal to

• A. $36$
• B. $55$
• C. $72$
• D. $45$

Answer 1

The correct option is A $36$

$x+y+z=10$

$x,y,z⟶$ positive

Number of Non negative for

$x,y,z⟶$ positive

$\Rightarrow x_1+x_2+x_3+............x_r=n{\Rightarrow }^{n+r-1}{C}_{r-1}x+y+z=10⟶for\Rightarrow a=x-1\Rightarrow b=y-1\Rightarrow c=z-1$

$a,b,c⟶$non negative integer

$\Rightarrow a+b+c=7\Rightarrow n=7\Rightarrow r=3$

Therefore number of possible solutions are

${\Rightarrow }^{7+3-1}{C}_{3-1}{\Rightarrow }^9{C}_2\Rightarrow \frac{9\times 8}{2}\Rightarrow 36$

Hence the correct answer is $36$