P´olya's Counting Theory - Home Page -- Tom Davis

where the c j corresponds to the j th chemical group.
Let’s look at the situation with different numbers of groups (colors), beginning with the simplest
situation: one group. Obviously, if all six positions are filled with the same group, there is only
one way to do it. In Equation 15, if there is one group, f i = c i 1, yielding:
P = c6 1 + 4c6 1 + 4c6 1 + 3c6 1 + 6c6 1
18
= 18c6 1
18 = c6 1 .
The coefficient 1 of c 6 1 means that there is exactly one way to construct an ethane derivative if all
six attached groups are the same.
Things get a bit more interesting with two different kinds of groups:
P = (c 1 + c 2 ) 6 + 4(c 1 + c 2 ) 3 (c 3 1 + c3 2 ) + 4(c3 1 + c3 2 )2 + 3(c 2 1 + c2 2 )3 + 6(c 6 1 + c6 2 )
18
= c 6 1 + c5 1 c 2 + 2c 4 1 c2 2 + 2c3 1 c3 2 + 2c2 1 c4 2 + c 1c 5 2 + c6 2
= c 6 1 + c6 2 +
c 5 1c 2 + c 1 c 5 2 +
2c 4 1 c2 2 + 2c2 1 c4 2 +
2c 3 1c 3 2.
Here is the expansion with three groups:
P =
(c 1 + c 2 + c 3 ) 6 + 4(c 1 + c 2 + c 3 ) 3 (c 3 1 + c 3 2 + c 3 3)+
4(c 3 1 + c3 2 + c3 3 )2 + 3(c 2 1 + c2 2 + c2 3 )3 + 6(c 6 1 + c6 2 + c6 3 )
18
= c 6 1 + c 5 1c 2 + c 5 1c 3 + 2c 4 1c 2 2 + 3c 4 1c 2 c 3 + 2c 4 1c 2 3 +
2c 3 1 c3 2 + 4c3 1 c2 2 c 3 + 4c 3 1 c 2c 2 3 + 2c3 1 c3 3 +
2c 2 1c 4 2 + 4c 2 1c 3 2c 3 + 6c 2 1c 2 2c 2 3 + 4c 2 1c 2 c 3 3 + 2c 2 1c 4 3 +
c 1 c 5 2 + 3c 1c 4 2 c 3 + 4c 1 c 3 2 c2 3 + 4c 1c 2 2 c3 3 + 3c 1c 2 c 4 3 + c 1c 5 3 +
c 6 2 + c 5 2c 3 + 2c 4 2c 2 3 + 2c 3 2c 3 3 + 2c 2 2c 4 3 + c 2 c 5 3 + c 6 3
= c 6 1 + c6 2 + c6 3 +
c 5 1c 2 + c 5 1c 3 + c 1 c 5 2 + c 1 c 5 3 + c 5 2c 3 + c 2 c 5 3 +
2c 4 1 c2 2 + 2c4 1 c2 3 + 2c2 1 c4 2 + 2c2 1 c4 3 + 2c4 2 c2 3 + 2c2 2 c4 3 +
4c 3 1c 2 2c 3 + 4c 3 1c 2 c 2 3 + 4c 2 1c 3 2c 3 + 4c 2 1c 2 c 3 3 + 4c 1 c 3 2c 2 3 + 4c 1 c 2 2c 3 3 +
3c 4 1 c 2c 3 + 3c 1 c 4 2 c 3 + 3c 1 c 2 c 4 3 +
2c 3 1 c3 2 + 2c3 1 c3 3 + 2c3 2 c3 3 +
6c 2 1c 2 2c 2 3
In each of the final two cases above, the product is rearranged to show that the coefficients on
“similar” expressions is the same. For example, in the expansion with three groups, every term
that consists of four copies of one color and two of another has the same coefficient of 2: 2c 4 1 c2 2 ,
2c 4 1c 2 3, 2c 2 1c 4 2, et cetera. This makes perfect sense, of course, since there should be the same number
25