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Topic:
number of ways to add and multiply numbers
Replies:
2
Last Post:
Aug 16, 2007 1:51 AM




Re: number of ways to add and multiply numbers
Posted:
Aug 16, 2007 1:51 AM


On Aug 15, 10:33 pm, stephane.fr...@gmail.com wrote: > I have an idea , I just wondered if i don't double count sometimes . > I consider k>2 components ( two operators + or * ) The number of > operations is N : > > N = 2*sum(i=0>k)C(k,i) + sum(i=0>k2)[sum(j=0>n)C(j,i)] > > Where C(k,i) is the binomial coefficient C(k,i)=k!(i!*(ki)!) > First term count the + or * only , the second term count the > permutation of * ( + are induite this way ) > > Hope this is good > > SF >  > **Apology if I'm wrong, It's only by trying that you can make mistakes > *** > > On Aug 15, 9:38 pm, "Jon Slaughter" <Jon_Slaugh...@Hotmail.com> wrote: > > > > > does anyone know the unique symbolic(and not computation) way to count the > > number of ways to add and multiply numbers from a set? > > > That is, say I'm given 1 and 2 then I can do > > > 1 > > 2 > > 1+2 > > 1*2 > > > for 1,2,3 > > > 1 > > 2 > > 3 > > 1+2 > > 1+3 > > 2+3 > > 1+2+3 > > 1*2 > > 1*3 > > 2*3 > > 1*2+3 > > 1+2*3 > > 1*3+2 > > 1*2*3 > > > (I think thats every one but I thought from previous work that there should > > be 15) > > > (1+3*2 is not in the set because the operations are commutative(its already > > there in the form of 1+2*3) > > > Actually the problem comes from trying to count the number of combinations > > of putting elements in a electrical network such as resistors. > > > Say I have a pile of n resistors then how many ways can I form different > > "networks" from them that result in unique results. Actually the problem is > > more general and one must treat each resistor as unique and that unique > > results do not relate to the electrical characteristics. > > > Basically you can form parallel or series networks recursively and there is > > a reduction in some cases in that if two elements are in series then there > > order doesn't matter(for my application) or if they are in parallel. But in > > some cases the order does matter because it cannot be reordered like the > > example with 1*3+2 > > > which would be > > > 1 > >  2 > > 3 > > > except for things like 2+1*3, 2+3*1, 3*1+2. > > > notice that for k elements there is k1 operands between them but the > > ordering is strange and I don't know how to describe it. > > > Any ideas? > > > Thanks, > > Jon Hide quoted text  > >  Show quoted text 
well i ve just realise the second term is wrong . You should read N = 2*sum(i=0>k)C(k,i) + sum(i=1>k2)[sum(j=3>n)C(j,i)]



