Sum of all possible products of elements taken from couples


by Wentu
Tags: combinatorics, couples, elements, products
Wentu
Wentu is offline
#1
Dec13-13, 12:11 PM
P: 10
Hello

I have N couples of real numbers higher than 1.
Let's call them like (a0,b0), (a1,b1),...,(aN,bN)
I have a number R <= N.

I need the sum of all the possible products of N elements, chosing one from each couple but exactly R times the "b" element and N-R times the "a" element.
Which is the best way to do it?

As an example:
(2,3), (5,7), (11,13)
N = 3, R = 2
I need 2x7x13 + 3x5x13 + 3x7x11

Thank you!
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1MileCrash
1MileCrash is online now
#2
Dec13-13, 12:45 PM
1MileCrash's Avatar
P: 1,227
As an expression I think what you want to do is:

[itex]\Sigma^{N}_{k=0} (a_{k}(\Sigma^{N}_{i=0} b_{i}))[/itex]

I have no idea if there is any way to compute this other than just doing it.

EDIT: Nevermind, I see you don't want "sum of all possible products of N+1 elements" but sum of all possible products of a choice of R elements from the N+1 elements. No idea, you're probably going to have to write a program for that.
Wentu
Wentu is offline
#3
Dec17-13, 01:24 AM
P: 10
I got the answer from "Michael":
It is the coefficient of x^R in (a0+xb0)(a1+xb1)...(aN+xbN)


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