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

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Solution

$(i)$ Since $x + y + z + w = 20$
Here $x \geq 0, y \geq 0, z \geq 0, w \geq 0$
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
$=^{20 + 4 - 1} C_{4 - 1}$
$=^{23} C_{3}$
$= \frac{23.22.21}{1.2.3} = 1771$
$(i i)$ Since $x + y + z + w = 20 . . . (i)$
Here $x \geq 1, y \geq 1, z \geq 1, w \geq 1$
or $x - 1 \geq 0, y - 1 \geq 0, z - 1 \geq 0, w - 1 \geq 0$
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 $(i)$
$x_{1} + 1 + y_{1} + 1 + z_{1} + 1 + w_{1} + 1 = 20$
$\Rightarrow x_{1} + y_{1} + z_{1} + w_{1} = 16$
and $x_{1} \geq 0, y_{1} \geq 0, z_{1} \geq 0, w_{1} \geq 0$
Hence the total number of solutions
$=^{17 + 4 - 1} C_{4 - 1}$
$=^{19} C_{3} = \frac{19.18.17}{1.2.3}$
$= 57. 17 = 969$
$Alternative Method:$
$(i i) ∵ x + y + z + w = 20$
$x \geq 1, y \geq 1, z \geq 1, w \geq 1$
Hence the total no. of solutions
$=^{20 - 1} C_{4 - 1} =^{19} C_{3} = 969$

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