Question

The number of non negative integral solutions of $x+2y+2z=21$ is

• A. 66
• B. 38
• C. 18
• D. 51

Answer 1

Answer: (A)

66

PROBLEMS BASED ON CERTAIN THEOREMS ON COMBINATIONS :

Number of non negative integral solutions of x 2 y 3z = n is coefficient of $x^n$ in ${(1-x)}^{-1}{(1-x^2)}^{-1}{(1-x^3)}^{-1}$

PROBLEMS BASED ON CERTAIN THEOREMS ON COMBINATIONS :

$\cdot$ The number of positive integral solutions of the equation $x_1+x_2+x_3+....+x_r=n$ is ${}^{n-1}{C}_{r-1}.$$\cdot$ The number of non-negative integral solutions of the equation<br>$x_1+x_2+x_3+....+x_r=n$ is ${}^{n+r-1}{C}_{r-1}.$

The given equation can be written as $x+2(x+y)=21$.
Hence, $x$ can take only odd values.
So, we will make the cases for $x$
a) When $x=1$ we have $y+z=10$ and the number of non-negative integer solutions for the equation are ${}^{10+2-1}{C}_{2-1}=11$
b) When When $x=3$ we have $y+z=9$ and the number of nonnegative integer solutions for the equation are ${}^{9+2-1}{C}_{2-1}=10$
c) When $x=5$ we have $y+z=8$ and the number of non-negative integer solutions for the equation are ${}^{8+2-1}{C}_{2-1}=9$
Continuing we get
When $x=21$ we have $y+z=0$ has only one solution.
Hence, the total number of solutions are $11+10+9+\cdots +1=66$