in Combinatory
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3 votes
3 votes

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\ $

in Combinatory
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4 Comments

@Satbir thanks for the reply..

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) .

 

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https://gateoverflow.in/blog/6412/generating-function-recurrence-relation-useful-link

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

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@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.
 

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