Find the number of non negative intagral solutions of $x+2y+z = 100 $. Given that $x, y, z \geq1$.

Question

I have taken the following approach using combinations.

I am considering $2y$ to be one object say $k$.

So now I will have the equation : $x+k+z = 100$

and the conditions : $x,z \geq1 $ and $k \geq2$.

Then in order to use the formula for distributing $n$ alike objects I convert this equation into

$X+K+Z =96$

where $X = x-1, K = k-2, Z = z-1$

and $X, Y, Z \geq0$

Now applying the formula for distribution of $n$ alike objects which is :

$\frac{(n+r-1)!} {n!(r-1)!} $

$\frac{(96+3-1)! }{96!(3-1)!} = 4753$

But the answer is supposed to be $2401$

Why is this method wrong and what is the right way for solving this problem?

Answer 1

The problem with your solution is that you assumed k can be anything but in fact k must be even. Here is my solution:

Consider the parity of $x$ and $z$.

if $x$ and $z$ are both odd: then let $x=2m+1, z=2n+1.$ Then $m+n\le48.$ There are $$49(m=0)+48(m=1)...+1(m=48)=1225$$ ways to choose $m+n$.

if $x$ and $z$ are both even: then let $x=2m, z=2n.$ Then $m+n\le49.$ There are $$ 48(m=1)+48(m=2)+...+1(m=48)=1176$$

Therefore the answer is $1225+1176=2401$

Answer 2

Modifying abc...'s answer:

Case $1$ ($x$ and $z$ odd), letting $x=2m-1$ and $z=2n-1$: $$2m+2y+2n=102\to m+y+n=51\to\binom{50}{2}=1225$$ Case $2$ ($x$ and $z$ even), letting $x=2m$ and $z=2n$: $$2m+2y+2n=100\to m+y+n=50\to\binom{49}{2}=1176$$ Total positive integer solutions:$$1225+1176=2401$$

Answer 3

Alternatively, when $y=1,2,3,...,49$, it is the sum: $$\sum_{y=1}^{49} {100-2y-1\choose 2-1}=\sum_{y=1}^{49} (99-2y)=99\cdot 49-2\cdot \frac{1+49}{2}\cdot 49=2401.$$