Question

Let the positive number of solutions of $x+y+z+w=20$ under the following conditions:
$\left(i\right)$ zero values of $x,y,z,w$ are included be "k"
$\left(ii\right)$ zero values are excluded be "m"
Find sum of digits of m+k ?

Answer 1

$\left(i\right)$ Since $x+y+z+w=20$
Here $x\ge 0,y\ge 0,z\ge 0,w\ge 0$
The number of solutions of the given equation in this case is same as the
number of ways of distributing $20$ things among $4$ different groups.
Hence total number of solutions
$={}^{20+4-1}{C}_{4-1}$
$={}^{23}{C}_3$
$=\frac{\mathrm{23.22.21}}{\mathrm{1.2.3}}=1771$
$\left(ii\right)$ Since $x+y+z+w=20...\left(i\right)$
Here $x\ge 1,y\ge 1,z\ge 1,w\ge 1$
or $x-1\ge 0,y-1\ge 0,z-1\ge 0,w-1\ge 0$
Let $x_1=x-1\Rightarrow x=x_1+1$
$y_1=y-1\Rightarrow y=y_1+1$
$z_1=z-1\Rightarrow z=z_1+1$
$w_1=w-1\Rightarrow w=w_1+1$
Then, from $\left(i\right)$
$x_1+1+y_1+1+z_1+1+w_1+1=20$
$\Rightarrow x_1+y_1+z_1+w_1=16$
and $x_1\ge 0,y_1\ge 0,z_1\ge 0,w_1\ge 0$
Hence the total number of solutions
$={}^{17+4-1}{C}_{4-1}$
$={}^{19}{C}_3=\frac{\mathrm{19.18.17}}{\mathrm{1.2.3}}$
$=57.17=969$
$\text{Alternative Method:}$
$\left(ii\right)\because x+y+z+w=20$
$x\ge 1,y\ge 1,z\ge 1,w\ge 1$
Hence the total no. of solutions
$={}^{20-1}{C}_{4-1}={}^{19}{C}_3=969$