Question

Check out this application of permutations and combinations in chemistry.

Answer 1

Consider 1,2,3 decatriene.\[CH2=C=C=CH-CH2-CH2-CH2-CH2-CH2-CH3\]
It can also be written as 1,2,4-decatriene, 1,4,7-decatriene, and many such combinations of the position of the three double bonds.
Calculate for a n-length hydrocarbon, the number of possible isomers(combinations of the positions of the double bonds) if it has r double bonds, if stereoisomers are not considered. That is, i dont want 1,2.3-deca triene, and 8,9,10-decatriene separately.

Answer 2

hmmmm... Interesting...
Lets see so in a n length carbon chain there n-1 bonds. Now let the bonds be at certain positions a,b,c...etc from one end. Now if x1,x2,x3......x r etc are number of bonds in between 2 double bonds.(Where r is number of double bonds). Now you can arrange x1,x2...xr in such a way that
x1+x2+x3+.....+xr=n-1-r.(r positions are already occupied so n-1-r are the bonds left).
This becomes the total number of non negative integral solutions of that equation.
You get (n-1-r+r-1 )C(r-1)= (n-2)C(r-1). But now it is symmetrical so i guess it is just 1/2 (n-2)C(r-1)

Answer 3

I meant let a,b,c are the double bonds Here a better figure for n=7 r=2 so then you need to find all possible solutions of x1+x2+x3=4.