I Discrete Optimization Problem?

Consider the expression:


$$A = \frac{ M! }{ r_1!\ r_2! }$$

where [itex] M = r_1 + r_2 [/itex],
where [itex] r_1 = (M - 2r_2) [/itex]

$$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 : [itex] r_2 = \frac{1}{2}r_1 [/itex].
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

Homework Helper
838
244
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

Homework Helper
838
244
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

Science Advisor
Homework Helper
10,698
1,697
Consider the expression:


$$A = \frac{ M! }{ r_1!\ r_2! }$$

where [itex] M = r_1 + r_2 [/itex],
where [itex] r_1 = (M - 2r_2) [/itex]

$$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 : [itex] r_2 = \frac{1}{2}r_1 [/itex].
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.##
 

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