How to find total no of integer solutions to the below equation ?

Question

In X+y+z<=12 , X,y,z>=1  

How to find total no of integral solutions ?If you are introducing any dummy variable for solving , please do explain the logic behind this .

Comments

dummy variable is added to satisfy equality.

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Yes  Ans:220

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For practice you can solve this question also

Determine the number of integer solutions for
$a)x_{1}+x_{2}+x_{3}+x_{4}+x_{5}<40$ where $x_{i}>=0$
$b)x_{1}+x_{2}+x_{3}+x_{4}+x_{5}<40$ where $x_{i}>=-3$

Ans

a) $44\choose  39$

b) $59\choose  54$
 

Answer 1

$x+y+x <= 12, \ \ \  \ \ \ \ \ \ \  x,y,z >0$  

to solve this problem with less steps we will introduce a dummy variable $w$ which will hold the value $12 - (x+y+z)$ so that RHS of equation becomes stable and we can solve this question easily.

so our new equation will be

$x + y  + z + w = 12, \ \ \ x,y,z >0$

now the constrain is $x,y,z>0$

so first we'll calculate all possible solutions then subtract the invalid solutions.

All possible solutions without any restriction = $\large \binom{15}{3}$ = $455$

Now calculating invalid solutions,

when $x$ is $0$,

$y+z+w = 12$

$= \large \binom{14}{2}$

 when $y$ is $0$,

$= \large \binom{14}{2}$

when $z$ is $0$,

$= \large \binom{14}{2}$

when both x and y are $0$,

z + w = 12

$= \large \binom{13}{1}$

similarly there are 2 more possibilities (x and z are 0 and z and y are 0),

when x,y and z (all 3) are 0,

$w = 12$

$= \large \binom{12}{0}$

So total invalid solutions = $ \large 3 \times \binom{14}{2} - 3 \binom{13}{1} + 1 \binom{12}{0}  = 235$

Total solutions = $455 - 235 =220 $

Comments

Yes Bro and you can also edit $X\geq1$

                                         $X-1\geq0$

                                      So,we can put $X_{1}=X-1$ $,X_{1}\geq0$

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No No

it's X1 = X+1

that's right

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No, brother

please check clearly