3

I have to find the number of solutions for: $$x_1 + x_2 + x_3 + x_4 = 42$$ when given: $$ (I) 12 <= x_1 <=13 $$ $$ (II) 3 <= x_2 <= 6 $$ $$ (III) 11 <= x_3 <= 18 $$ $$ (IV) 6 <= x_4 <= 10 $$

What I did so far, is define: $$ y_1 = x_1 - 12 $$ $$ y_2 = x_2 - 3 $$ $$ y_3 = x_3 - 11 $$ $$ y_4 = x_4 - 6 $$

Thus we get: $$ y_1 + y_2 + y_3 + y_4 = 10 $$ where $$ 0 <= y_1 <= 1 $$ $$ 0 <= y_2 <= 3 $$ $$ 0 <= y_3 <= 7 $$ $$ 0 <= y_4 <= 4 $$

now I defined: $\text{S = all solutions to}$ $y_1 + y_2 + y_3 + y_4 = 20$ $\text{where all}$ $y_i >=0$ $\text{without the upper limitation}$

and I know I have to subtract the rest of the cases, when $y_1 is >1$ and $y_2 is >3$ etc... but how do I count all those side cases that I should subtract?

tryst with freedom
  • 11,538
TheNotMe
  • 4,841

1 Answers1

1

For the newer information on question, since $x_i$ are integral(which is same as saying that they are integers):

The answer will be coefficient of $x^{10}$ in the following expansion:

$$ \begin{align} f(x) &= (x^0 + x^1)(x^0 + x^1 + x^2 + x^3)(x^0 + x^1 + x^2 + x^3 + x^4 + x^5 + x^6 + x^7)(x^0 + x^1 + x^2 + x^3 + x^4) \\ &= (1 + x)(1 + x + x^2 + x^3)(1 + x + x^2 + x^3 + x^4)(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7) \end{align} $$

hjpotter92
  • 3,049
  • im sorry. original is =42. my mistake – TheNotMe Apr 09 '13 at 05:38
  • What if I give y1 "2" because I know it is at least 2, then calculate A1 (the number of solutions when y1 > 1) and etc? would that solve it and give me the size of each Ai? – TheNotMe Apr 09 '13 at 18:14
  • you may have interests in this too: http://math.stackexchange.com/questions/751167/counting-problem-very-interesting-modular-n-algebraic-eqs-for-combinatorics?lq=1 – annie marie cœur Apr 12 '14 at 22:38