Find the number of whole number solutions for this equation
$ x_1 + x_2 + x_3 + x_4 = 11 $
where
$ x_1 \ge -2, x_2 \ge -1, x_3 \ge 0, x_4 \ge 5 $
Asked
Active
Viewed 176 times
2
-
1you must do case work! – Dr. Sonnhard Graubner Oct 12 '16 at 16:16
-
@Dr.SonnhardGraubner What do you mean? – Smebbs Oct 12 '16 at 16:17
-
begin wth the conditon $$x_4\geq 5$$ – Dr. Sonnhard Graubner Oct 12 '16 at 16:21
2 Answers
5
Write $x_1 = -2 + y_1, x_2 = -1 + y_1, x_3=y_3 x_4 = 5 + y_4$. Note that then $y_1,y_2,y_3,y_4 \ge 0$ and after substituting you will have:
$$y_1 + y_2 + y_3 + y_4 = 9$$
You can find all non-negative solution to this equation using Stars and Bars.
Stefan4024
- 35,843
4
*$ x_1 + x_2 + x_3 + ...+x_k = n \to \left(\begin{array}{c}n+k-1\\ k-1\end{array}\right) $
where
$ x_1 \ge -2, x_2 \ge -1, x_3 \ge 0, x_4 \ge 5 $ $$ x_1 \ge -2 \to y_1=x_1 +2\ge 0\\ x_2 \ge -1 \to y_2=x_2 +1\ge 0\\ x_3 \ge 0\\ x_4 \ge 5 \to y_4=x_4 -5\ge 0$$ $$\color{red} {x_1+x_2+x_3+x_4=11\\(x_1+2)+(x_2+1)+(x_3)+(x_4-5)=11+2+1-5=9\\y_1+y_2+x_3+y_4=9} \to \left(\begin{array}{c}9+4-1\\ 4-1\end{array}\right)=\\\left(\begin{array}{c}12\\ 3\end{array}\right)=\frac{12.11.10}{3.2.1}=220$$
Khosrotash
- 24,922
-
the first equation meaning that you are to find the number of weak compositions of $n$ into $k$ parts – G Cab Oct 12 '16 at 17:02