Question

Let the positive number of solutions of x +y +z + w = 20 under the following condition.
(1) Zero values of $x,y,z,w$ are included to be “k”.
(2) Zero values are excluded by “m”. Find the sum of digits of m + k.

Answer 1

Hint: Here we solve the problem by using the formula of finding the number of solutions for any linear equation. Therefore the formula to find the number of solutions for an equation is $^{n + r - 1}{C_{r - 1}}$, where n is the positive constant on the right hand side of the equation and r is the numbers of variables in the equation.

Complete step by step answer:
(1) Given equation is $x + y + z + w = 20$
Here if we observe the given equation we can say that $x \geqslant 0,y \geqslant 0,z \geqslant 0,w \geqslant 0$.
If we observe the first part of the question we can assume that there are 20 similar things which are distributed among 4 different groups.
We know that the number of solutions of a given equation in this case is the same as the number of ways of distributing 20 similar things among 4 different groups.
And in the question, the number of solutions for this given equation is k. So we have:
$ \Rightarrow k{ = ^{20 + 4 - 1}}{C_{4 - 1}}$
$ \Rightarrow k{ = ^{23}}{C_3}$
We know that $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, using this we’ll get:
$
   \Rightarrow k = \dfrac{{23!}}{{3! \times 20!}} \\
   \Rightarrow k = \dfrac{{23 \times 22 \times 21}}{{3 \times 2 \times 1}} = 1771 \\
$
Thus according to the question, $k = 1771$

(2) Since $x + y + z + w = 20 - - - - - - > (1)$
Here according to second part of the question we can say that $x \geqslant 1,y \geqslant 1,z \geqslant 1,w \geqslant 1$ or
$x - 1 \geqslant 0,y - 1 \geqslant 0,z - 1 \geqslant ,w - 1 \geqslant 0$
Now here let us consider that
${x_1} = x - 1 \Rightarrow x = {x_1} + 1$
${y_1} = y - 1 \Rightarrow y = {y_1} + 1$
${z_1} = z - 1 \Rightarrow z = {z_1} + 1$
${w_1} = w - 1 \Rightarrow w = {w_1} + 1$
Then by using above values we can write equation (1) as
$ \Rightarrow {x_1} + 1 + {y_1} + 1 + {z_1} + 1 + {w_1} + 1 = 20$
$
   \Rightarrow {x_1} + {y_1} + {z_1} + {w_1} = 20 - 4 \\
   \Rightarrow {x_1} + {y_1} + {z_1} + {w_1} = 16 \\
$
And here we know that ${x_1} \geqslant 1,{y_1} \geqslant 1,{z_1} \geqslant 1,{w_1} \geqslant 1$.
From the question, the number of solutions for this given equation is m. So we have:
$
   \Rightarrow m{ = ^{16 + 4 - 1}}{C_{4 - 1}} \\
   \Rightarrow m{ = ^{19}}{C_3} \\
$
Again using $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we’ll get:
$
   \Rightarrow m = \dfrac{{19!}}{{3! \times 16!}} \\
   \Rightarrow m = \dfrac{{19 \times 18 \times 17}}{{3 \times 2}}, \\
   \Rightarrow m = 969 \\
$
Now we have the values of m and k. Thus:
$
   \Rightarrow m + k = 969 + 1771 \\
   \Rightarrow m + k = 2740 \\
$
Sum of digits of $m + k$ $ = 2 + 7 + 4 + 0 = 13$

NOTE: The above used method is for finding the number of ways of distributing n similar objects into k different groups. If the objects are also different from each other then this approach is not applicable.