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Question: The total number of positive integral solution of 15 < \[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}\le 20\] is eq...

The total number of positive integral solution of 15 < x1+x2+x3≤20{{x}_{1}}+{{x}_{2}}+{{x}_{3}}\le 20 is equal to
(a) 685
(b) 785
(c) 1125
(d) None of these

Explanation

Solution

Hint: According to question, we have to find the number of ways in which the sum of x1,x2,{{x}_{1}},{{x}_{2}}, and x3{{x}_{3}} is equal to a particular number. These particular numbers are 16,17,18,19 and 20. So, for solving this question, we will use a partition method. This method says that the number of ways of dividing ‘n’ identical object into ‘r’ distinct things where each thing can get any number of objects, is given by n+r−1Cr−1^{n+r-1}{{C}_{r-1}}. So, now we are in condition to solve this question.

Complete step-by-step answer:
In the first step, let us take the sum of x1,x2{{x}_{1}},{{x}_{_{2}}} and x3{{x}_{3}} be equal to 16. Thus,
x1+x2+x3{{x}_{1}}+{{x}_{_{2}}}+{{x}_{3}}=16…… (1)
As x1,x2{{x}_{1}},{{x}_{_{2}}} and x3{{x}_{3}} are positive integers, the minimum value of x1,x2{{x}_{1}},{{x}_{_{2}}} and x3{{x}_{3}} are 1 each, we assume x1=t1+1,x2=t2+1,{{x}_{1}}={{t}_{1}}+1,{{x}_{2}}={{t}_{2}}+1, andx3=t3+1{{x}_{3}}={{t}_{3}}+1. So the equation (i) becomes: {{t}_{1}}+$$$${{t}_{2}}$$$$+{{t}_{3}}=13. We can solve this equation by partition method. So the number of ways in which we can obtain the above equation is given by the formula: n+r−1Cr−1^{n+r-1}{{C}_{r-1}}, where n=13 and r=3. So the total number of ways becomes,

13+3−1C3−1=15C2=105^{13+3-1}{{C}_{3-1}}{{=}^{15}}{{C}_{2}}=105
In the second case, x1+x2+x3{{x}_{1}}+{{x}_{_{2}}}+{{x}_{3}}=17. Again we will convert it into {{t}_{1}}+$$$${{t}_{2}}$$$$+{{t}_{3}}=14. Number of ways in obtaining the above equation is given by n+r−1Cr−1^{n+r-1}{{C}_{r-1}}=14+3−1C3−1=16C2=120^{14+3-1}{{C}_{3-1}}{{=}^{16}}{{C}_{2}}=120
Where n=14 and r=3 respectively. We can do similarly for the rest of the cases, for example, when the sum of thex1,x2{{x}_{1}},{{x}_{2}} and x3{{x}_{3}} will be equal to 18, then the total number of ways will be17C2^{17}{{C}_{2}}. Similarly, in case of 19 and 20, the total number of ways to obtain the sum will be equal to 18C2^{18}{{C}_{2}}and 19C2^{19}{{C}_{2}} respectively. So the total number of ways can be obtained by adding the number of ways to calculate the sum of 16,17,18,19 and 20 together. So the total number of positive integral solution is obtained by: -
Total= 15C2^{15}{{C}_{2}}+16C2^{16}{{C}_{2}}+17C2^{17}{{C}_{2}}+18C2^{18}{{C}_{2}}+19C2^{19}{{C}_{2}}
=105+120+136+153+171
=685
Hence, the total number of positive integral solutions of 15< x1+x2+x3≤20{{x}_{1}}+{{x}_{2}}+{{x}_{3}}\le 20 is 685.
Hence, option (a) is correct.

Note: we cannot use partition method directly in the solution because the condition for the partition method is that any number of objects can be distributed to any number of things but in the above question, we have to find only the positive solution that is, minimum value of x1,x2{{x}_{1}},{{x}_{_{2}}} and x3{{x}_{3}}are 1 each. That’s why we have assigned x1=t1+1,x2=t2+1,{{x}_{1}}={{t}_{1}}+1,{{x}_{2}}={{t}_{2}}+1, and x3=t3+1{{x}_{3}}={{t}_{3}}+1(where t1,t2{{t}_{1,}}{{t}_{2}}and t3{{t}_{3}} belongs to whole number) to meet the condition of partition method.