Discrete Optimization Problem?

[iScience]

Consider the expression:$$A = \frac{ M! }{ r_1!\ r_2! }$$

where $M = r_1 + r_2$,
where $r_1 = (M - 2r_2)$

$$A = \frac{ (r_1 + r_2)! }{ r_1!\ r_2! } \\ \ \\ \
= \frac{ ((M-2r_2) + r_2)! }{ (M-2r_2)!\ (r_2)! } \\ \ \\ \
= \frac{ (M-r_2)! }{ (M-2r_2)!\ r_2! }

$$

Then, for a given M, A maximum occurs at : $r_2 = \frac{1}{2}r_1$.
I know this because I can generate a table algorithmically. But I'd like to know how to arrive here analytically. Any pointers would be appreciated.

 

[tnich]

Given that $M = r_1 + r_2$ and $r_1 = (M - 2r_2)$, then if we substitute the second expression for $M$ in the first expression, we get $M = M - r_2$, implying that $r_2 = 0$. Then $A ≡ 1$.

 

[tnich]

Given that $M = r_1 + r_2$ and $r_1 = (M - 2r_2)$, then if we substitute the second expression for $M$ in the first expression, we get $M = M - r_2$, implying that $r_2 = 0$. Then $A ≡ 1$.

Maybe you meant to say $r_1 = M - r_2$

 

[Ray Vickson]

Consider the expression:$$A = \frac{ M! }{ r_1!\ r_2! }$$

where $M = r_1 + r_2$,
where $r_1 = (M - 2r_2)$

$$A = \frac{ (r_1 + r_2)! }{ r_1!\ r_2! } \\ \ \\ \
= \frac{ ((M-2r_2) + r_2)! }{ (M-2r_2)!\ (r_2)! } \\ \ \\ \
= \frac{ (M-r_2)! }{ (M-2r_2)!\ r_2! }

$$

Then, for a given M, A maximum occurs at : $r_2 = \frac{1}{2}r_1$.
I know this because I can generate a table algorithmically. But I'd like to know how to arrive here analytically. Any pointers would be appreciated.

For
$$ A(r) = \frac{M!}{r! (M-r)!},$$
look at the ratio
$$R(r) = \frac{A(r+1)}{A(r)}$$
and examine when $R(r)$ goes from $R>1$ to $R < 1$ as $r$ increases.

BTW: having $r_1 = \frac{M}{2}$ is possible only if $M$ is even; if $M$ is odd, the solution will be at the two neighboring integers to $M/2.$