2 votes

positive integral solution should be

(1,1,1)

(1,1,2)

(1,2,1)

(1,3,1)

(1,1,3)

(1,2,2)

(1,2,3)

(1,3,2)

(1,4,1)

(1,1,4)

(1,1,1)

(1,1,2)

(1,2,1)

(1,3,1)

(1,1,3)

(1,2,2)

(1,2,3)

(1,3,2)

(1,4,1)

(1,1,4)

2 votes

Given x1,x2,x3 all are positive integers => x1,x2,x3 >=1

Now we can write 15x1+x2+x3 <= 20 as x2 +x3 <= 20 - 15x1.

Now if x1=1, x2+x3 <= 20-15 = 5 => if x1= 1, x2+x3 <= 5.

for x1 >= 2, x2+x3 <= 20-30 = -10 which is not possible because x2,x3 are positive which can't give -ve on adding together.

So only possibility is : x1 = 1 & x2+x3 <= 5;

Now solving x2 + x3 <= 5,

if x2 =1, x3 <= 5-1 = 4 => for x2 =1, we have x3 = 1,2,3,4

similarly for x2=2, x3 <= 5-2 = 3 => for x2 =2, we have x3 = 1,2,3

similarly for x2=3, x3 <= 5-3 = 2 => for x2 =3, we have x3 = 1,2

similarly for x2=4, x3 <= 5-4 = 1 => for x2 =4, we have x3 = 1

x2 >= 5 not possible as it gives '0' or -ve value for x3 since x3 is +ve integer.

So we have x1=1, x2 =1, x3 = 1,2,3,4 => 4 solution

x1=1, x2 =2, x3 = 1,2,3 => 3 solution

x1=1, x2 =3, x3 = 1,2 => 2 solution

x1=1, x2 =4, x3 = 1=> 1 solution

Hence Total no. of solutions are = 4+3+2+1 =10

**Alternate Method:**

Given 15x1 + x2 +x3 <=20 , Here x1,x2,x3 >=1

Rewriting above inequation as follows: 15(x1-1) + (x2-1) + (x3-1) <= 20 - (15+1+1) = 3

So inequality reduces to : 15(x1-1) + (x2-1) + (x3-1) <= 3

Let a = x1-1; b= x2-1; c= x3-1; So we have 15a+b+c <=3 ,a,b,c >=0

Let us add d>=0 on LHS so that we have 15a+b+c+d = 3 where a,b,c,d >=0.

So no. of +ve integral solutions of 15x1+x2+x3 <= 20 is equivalent to no. of non-negative solutions for 15a+b+c+d = 3 where a,b,c,d >=0.

Hence reqd no of solution = coefficient of x^{3 }in (1+x^{15 }+x^{30} +x^{45})(1+x^{1}+x^{2} +x^{3})^{3}

Ignoring higher coeff.(>=15) in the expression as it doesn't give reqd coeff.

we have, reqd no of solution = coefficient of x^{3 }in (1)(1+x^{1}+x^{2} +x^{3})^{3}

^{ = } ^{3+3-1}C_{3 } = ^{5}C_{3 }=10 (Used standard formula for finding coeff.)

Hence reqd. no. of solutions = 10.