Question

Find the total number of probable values of the equation $x + y + x + w = 20$ (including zero) without restrictions?
(a) $1671$
(b) $1770$
(c) $1771$
(d) $1670$

Answer 1

Hint: Here, basically we are going to analyze or say, assume the each terms x, y, z, and w respectively greater than zero as the question implies to solve the output including zero. And, then making these groups and substituting in the combinatorial equation that is $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$the desired value is obtained.

Complete step-by-step solution:
The given equation implies that,
$x + y + x + w = 20$
But, the characteristics given condition clarifies including zero and without the restrictions of any number
As a result, we can write that
$x \geqslant 0$,$y \geqslant 0$,$z \geqslant 0$and$w \geqslant 0$
Since, there exists four different functions namely x, y, z, and w respectively
Therefore, dividing the respective solution in total four different groups which seems that the number of required solutions of the given equations remains equal to the particular system of calculations,
Hence, using the permutations and combinations concept of logic, we get
$ {\text{ }}{ = ^{20 + 4 - 1}}{C_{4 - 1}}$
Where, $20 - 4 - 1$implies the combination total number of groups divided in to the four parts having single divisible (say, x, y, z, and w each respectively)
(Which is in compare with factorial terminology $^n{C_r}$where ‘n ‘is natural or real parameter in the question and ‘r’ is the difference term in the sequence, )
The equation becomes,
$ {\text{ }}{{\text{ = }}^{23}}{{\text{C}}_3}$
Using the combinational statement that is $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$, we get
${ \Rightarrow ^{23}}{C_3} = \dfrac{{23!}}{{3!(23 - 3)!}}$
Simplifying the above equation predominantly, we get
${ \Rightarrow ^{23}}{C_3} = \dfrac{{23!}}{{3! \times 20!}}$
Now , by the factorial definition it seems that,
${ \Rightarrow ^{23}}{C_3} = \dfrac{{23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{(3 \times 2 \times 1) \times (20 \times 19 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}}$
Solving the above equation mathematically, we get
$
  { \Rightarrow ^{23}}{C_3} = \dfrac{{25852016738884976640000}}{{14597412049059840000}} \\
  { \Rightarrow ^{23}}{C_3} = 1771 \\
 $
$\therefore $ The option (c) is absolutely correct!

Note: One must remember the definition of factorial so as to clear the concept behind the problem or intention asked to solve. Basically, factorial is the multiplication of the number below it. As a result, to a desired value the group or difference need to be analyzed properly with the proper subset of the problem. Also, the combinational statement needs to be mug-up in sequence i.e. $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$particularly.