Question

The total number of positive integral solution of $15 < {x_1} + {x_2} + {x_3} \leqslant 20$ is equal to
A. 685
B. 785
C. 1125
D. None of these

Answer 1

Here, we will first assume that the sum ${x_1} + {x_2} + {x_3}$ is $16 + r$, where $r$ is 0,1,2,3,4. Then we will take ${y_1} = {x_1} + 1$, ${y_2} + 1 = {x_2}$ and ${y_3} + 1 = {x_3}$ in the above equation after the required combinations, we will use the property of combinations, that is, ${}^n{C_r} = {}^n{C_{n - r}}$, where $n$ is the total numbers and $r$ is the required numbers. And then take the given conditions of the question, to find the required value.

Complete step by step answer:

We are given that the inequality is $15 < {x_1} + {x_2} + {x_3} \leqslant 20$.
We have seen that the above the sum ${x_1} + {x_2} + {x_3}$ is greater than 15 and less than or equal to 20.
Let us assume that the sum ${x_1} + {x_2} + {x_3}$ is $16 + r$, where $r$ is 0,1,2,3,4.
Then we have ${x_1} + {x_2} + {x_3} = 16 + r$, where ${x_1} \geqslant 0$, ${x_2} \geqslant 0$, ${x_3} \geqslant 0$.
Putting ${y_1} = {x_1} + 1$, ${y_2} + 1 = {x_2}$ and ${y_3} + 1 = {x_3}$ in the above equation, we get

$$\Rightarrow {y_1} + 1 + {y_2} + 1 + {y_3} + 1 = 16 + r \\\ \Rightarrow {y_1} + {y_2} + {y_3} + 3 = 16 + r \\\ \Rightarrow {y_1} + {y_2} + {y_3} = 13 + r \\\$$

Now we will find the number of positive integral solutions using the combinations of ${y_1} + {y_2} + {y_3} = 13 + r$, where ${y_1} \geqslant 0$, ${y_2} \geqslant 0$, ${y_3} \geqslant 0$.
$\Rightarrow {}^{13 + r + 3 - 1}{C_{13 + r}}$
Simplifying the above expression, we get

$$\Rightarrow {}^{13 + r + 2}{C_{13 + r}} \\\ \Rightarrow {}^{15 + r}{C_{13 + r}} \\\$$

Using the property of combinations, that is,${}^n{C_r} = {}^n{C_{n - r}}$, where $n$ is the total numbers and $r$ is the required numbers in the above expression, we get

$$\Rightarrow {}^{15 + r}{C_{15 + r - 13 - r}} \\\ \Rightarrow {}^{15 + r}{C_2} \\\$$

Thus, the total number of solutions is $\sum\limits_{r = 0}^4 {{}^{15 + r}{C_2}}$.
Simplifying the obtained expression for the total number of solutions, we get

$$\Rightarrow {}^{15 + 0}{C_2} + {}^{15 + 1}{C_2} + {}^{15 + 2}{C_2} + {}^{15 + 3}{C_2} + {}^{15 + 4}{C_2} \\\ \Rightarrow {}^{15}{C_2} + {}^{16}{C_2} + {}^{17}{C_2} + {}^{18}{C_2} + {}^{19}{C_2} \\\$$

We know that the formula of solving combinations is ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, where $n$ is the number of items, and $r$ is the number of item being chosen at a time.
Calculating the combinations using the above formula in the above expression, we get

$$\Rightarrow \dfrac{{15!}}{{2!\left( {15 - 2} \right)!}} + \dfrac{{16!}}{{2!\left( {16 - 2} \right)!}} + \dfrac{{17!}}{{2!\left( {17 - 2} \right)!}} + \dfrac{{18!}}{{2!\left( {18 - 2} \right)!}} + \dfrac{{19!}}{{2!\left( {19 - 2} \right)!}} \\\ \Rightarrow \dfrac{{15!}}{{2!13!}} + \dfrac{{16!}}{{2!15!}} + \dfrac{{17!}}{{2!15!}} + \dfrac{{18!}}{{2!16!}} + \dfrac{{19!}}{{2!17!}} \\\$$

Simplifying the factorials in the above expression, we get
$\Rightarrow \dfrac{{15 \times 14 \times 13!}}{{2!13!}} + \dfrac{{16 \times 15 \times 14!}}{{2!14!}} + \dfrac{{17 \times 16 \times 15!}}{{2!15!}} + \dfrac{{18 \times 17 \times 16!}}{{2!16!}} + \dfrac{{19 \times 17 \times 18!}}{{2!18!}}$
We will now cancel the same factorials in numerators and denominators in the above expression, we get
$\Rightarrow \dfrac{{15 \times 14}}{2} + \dfrac{{16 \times 15}}{2} + \dfrac{{17 \times 16}}{2} + \dfrac{{18 \times 17}}{2} + \dfrac{{19 \times 18}}{2}$
Taking $\dfrac{1}{2}$ common in the above expression, we get

$$\Rightarrow \dfrac{1}{2}\left( {15 \times 14 + 16 \times 15 + 17 \times 16 + 18 \times 17 + 18 \times 19} \right) \\\ \Rightarrow \dfrac{1}{2}\left( {210 + 240 + 272 + 306 + 342} \right) \\\ \Rightarrow \dfrac{1}{2}\left( {1370} \right) \\\ \Rightarrow 685 \\\$$

Therefore, the total number of positive integral solution of $15 < {x_1} + {x_2} + {x_3} \leqslant 20$ is equal to 685.
Hence, option A is correct.

Note: In solving these types of questions, you should be familiar with the concept of combinations, their simplification, and factorial distribution. While counting the possible combination factors in the above question, the factors could be chosen such that the factor on the left is always greater than the factor on the right, never less. This will help in rectifying the redundant cases. Some students mistakenly use the formula of permutations, ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ instead of the formula of combinations, so one should kept in mind such formulas while solving problems like this.