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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)
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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 xin (1+x15 +x30 +x45)(1+x1+x2 +x3)3

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

we have, reqd no of solution =  coefficient of xin (1)(1+x1+x2 +x3)3

 =  3+3-1C3   = 5C3  =10   (Used standard formula for finding coeff.)

Hence reqd.  no. of solutions =  10.

edited by

4 Comments

yup it is from jnu papers,.
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k...options seem wrong.Though i have given two methods for solving this qn,i can put more alternate method to solve same qn...and all give same ans.
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u r correct shashank kumar . thier is a mistake in options
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0

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