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Find the number of possible solutions for $x,y,z$ for each the following cases.

$Case\ 1.$    Case of unlimited repetition.

$x + y +z = 10$ and $x \geq 0\ , y \geq 0,\ z \geq 0$

$Case\ 2$   Case of unlimited repetition with variable lower bounds

$x + y +z = 10$ and $x \geq 1\ , y \geq 2 \ , z \geq 3\$

$Case\ 3$    case when upper bounds and lower bounds are given , no variable can be greater than upper bound.

$x + y +z = 10$ and $10 \geq x \geq 0\ , \ 10 \geq y \geq 0 \ , 10 \geq z \geq 0\$

$Case\ 4$    case when upper bounds and lower bounds are given , lower bounds are different, no variable can be greater than upper bound

$x + y +z = 10$ and $10 \geq x \geq 1\ , \ 10 \geq y \geq 2 \ , 10 \geq z \geq 3\$

$Case\ 5$    case when upper bounds and lower bounds are given , only 1 variable can be greater than upper bound in the invalid case that gives correct sum like 11+2+3 = 16

$x + y +z = 16$ and $10 \geq x \geq 0\ , \ 10 \geq y \geq 0 \ , 10 \geq z \geq 0\$

$Case\ 6$    case when upper bounds and lower bounds are given , lower bounds are different ,only 1 variable can be greater than upper bound in the invalid case that gives correct sum like 11+2+3 = 16

$x + y +z = 16$ and $10 \geq x \geq 1\ , \ 10 \geq y \geq 2 \ , 10 \geq z \geq 3\$

$Case\ 7$    case when upper bounds and lower bounds are given , only 2 variable can be greater than upper bound in the invalid case that gives correct sum  like 11+11+3 =25

$x + y +z = 25$ and $10 \geq x \geq 0\ , \ 10 \geq y \geq 0 \ , 10 \geq z \geq 0\$

$Case\ 8$    case when upper bounds and lower bounds are given , lower bounds are different ,only 2 variable can be greater than upper bound in the invalid case that gives correct sum like 11+11+3 =25

$x + y +z = 25$ and $10 \geq x \geq 1\ , \ 10 \geq y \geq 2 \ , 10 \geq z \geq 3\$

$Case\ 9$   case when upper bounds and lower bounds are given , lower bounds are different ,all variable can be greater than upper bound in the invalid case that gives correct sum like 11+11+11 = 33

$x + y +z = 33$ and $10 \geq x \geq 1\ , \ 10 \geq y \geq 2 \ , 10 \geq z \geq 3\$

$Case\ 10$    case when upper bounds and lower bounds are given , and both upper bound and lower bounds are variable.

$x + y +z = 10$ and $8 \geq x \geq 1\ , \ 20 \geq y \geq 2 \ , 12 \geq z \geq 3\$

i have one more question. is there any good way to calculate coefficient of $a^{s}$ in the generating function in this comment..(https://gateoverflow.in/325255/the-interesting-combination-sum-problems?show=325256#c325256) .

Try to see questions on generating function on GO they would help.

@chirudeepnamini

It's just a variable in the polynomial expression. Combinatorial problems can be argued using the multiplication of polynomial expressions. BTW you can watch this video to grasp the basic idea.