# Math Help - Ho many integral solutions are there...

1. ## Ho many integral solutions are there...

I am asked to find how many integral solutions there..

x1+x2+x3+..........+x10=27

a.if all x>=1

b.if x1 ≥ 2, x2 ≥ 3, and the remaining xi ≥ 0

c.if all xi ≥ 1, and x1 ≤ 16

d.if all xi ≥ 1, and x1 ≤ 3 and x2 ≤ 2

now for the first one it was easy I think

26 choose 9

But once it is >=0 I get a little confused.

First say it said "all Xi>=0"

Is the formula "n+k-1 choose k"?

Then for question b. would I go...

Y1=X1-1
Y2=X2-2

Then +1 for each of the remaining xi which is +8?

So for b it would be

27-1-2+8=32

So...

b. 32 choose 9?

For
c.

Would it be 26 choose 9 subtract 10 choose 9 (number of solution with x1>16)

2. ## Re: Ho many integral solutions are there...

The $x_i\geq 0$ is just a condition that needs to be satisfied in order for a finite amount of solutions to exist. If anything other than positive integer solutions were possible among the x values, there would be infinitely many solutions to the equation. Think about it, say the situation in which $x_3 = 1$ and $x_4 = -1$ satisfied the equation, then so would $x_3 = 2$ and $x_4 = -2$, $x_3 = 3$ and $x_4 = -3$ etc.

b) $(27-(2+3)+10-1)\choose (10-1)$= ${31}\choose{9}$

3. ## Re: Ho many integral solutions are there...

Originally Posted by ehpoc
I am asked to find how many integral solutions there.
x1+x2+x3+..........+x10=27

c.if all xi ≥ 1, and x1 ≤ 16

d.if all xi ≥ 1, and x1 ≤ 3 and x2 ≤ 2
Is the formula "n+k-1 choose k"?

For
c.
Would it be 26 choose 9 subtract 10 choose 9 (number of solution with x1>16)
A bit LaTeX makes it easier to read.
[TEX]\binom{26}{9}[/TEX] gives $\binom{26}{9}$
[TEX]x_1+x_2+\cdots+x_{10}=27[/TEX] gives $x_1+x_2+\cdots+x_{10}=27$.

For c) note that there is only ten cases where $x_1>16$ if $x_i\ge 1$. Why?