Use three generating functions ($A, B$ and $C$) in three variables
($z$, $w$, and $v$) to enumerate strings ending in one of three colors
with the variable $u$ marking pairs of distinct adjacent colors. We
have $a$ instances of the first color, $b$ of the second and $c$ of
the third.
We obtain
$$A - z = Az + Buz + Cuz,
\\ B - w = Auw + Bw + Cuw,
\\ C - v = Auv + Buv + Cv.$$
Putting $Q = A + B + C$ we seek
$$P= \left. \frac{d}{du} Q \right|_{u=1}.$$
Differentiate to obtain
$$A' = A'z + Bz + B'uz + Cz + C'uz.
\\ B' = Aw + A'uw + B'w + Cw + C'uw,
\\ C' = Av + A'uv + Bv + B'uv + C'v.$$
Add these and put $u=1$ to get
$$P = P (z+w+v) +
\left.Q (z+w+v)\right|_{u=1}
- \left.(Az + Bw + Cv)\right|_{u=1}.$$
We have by inspection that
$$\left.Q \right|_{u=1} = \frac{z+w+v}{1-z-w-v}$$
and
$$\left.(Az + Bw + Cv)\right|_{u=1}
= \frac{z^2+w^2+v^2}{1-z-w-v}.$$
This yields
$$P(1-z-w-v) = \frac{(z+w+v)^2}{1-z-w-v}
- \frac{z^2+w^2+v^2}{1-z-w-v}$$
or
$$P = \frac{2zw+2zv+2wv}{(1-z-w-v)^2}.$$
We now skip ahead and show how to solve the general case of $m$
colors. We get for the generating function
$$P = \frac{2\sum_{1\le p \lt q\le m} w_p w_q}
{\left(1-\sum_{p=1}^m w_p\right)^2}.$$
Extracting coefficients on $[w_1^{d_1} w_2^{d_2}\cdots w_m^{d_m}]$
where $d = \sum_{p=1}^m d_p$ we use the Newton binomial and
obtain
$$2 \sum_{1\le p \lt q\le m}
(d-1) {d-2\choose d_1, d_2, \ldots d_p-1, \ldots d_q-1, \ldots d_m}
\\ = {d\choose d_1, d_2, \ldots d_p, \ldots d_q, \ldots d_m}
\\ \times 2 \sum_{1\le p \lt q\le m} (d-1)
\frac{1}{d(d-1)} d_p d_q.$$
Divide by the multinomial coefficient to obtain the expectation
$$\bbox[5px,border:2px solid #00A000]{
\frac{2}{d_1+d_2+\cdots+d_m}
\sum_{1\le p \lt q\le m} d_p d_q.}$$
The original problem by the OP then produces
$$\bbox[5px,border:2px solid #00A000]{
\frac{2ab+2ac+2bc}{a+b+c}.}$$
In particular we get for $(6,5,4)$ the exact value and the numerics
$$\bbox[5px,border:2px solid #00A000]{
\frac{148}{15} \approx 9.866666667.}$$
There is a Maple script for this which goes as follows (warning:
enumeration -- use on small values):
with(combinat);
ENUM :=
proc(L)
option remember;
local m, d, all, perm, pos, res, src, flips;
m := nops(L); d := add(p, p in L);
all := 0; res := 0;
src := [seq(seq(p, q=1..L[p]), p=1..m)];
for perm in permute(src) do
flips := 0;
for pos to d-1 do
if perm[pos] <> perm[pos+1] then
flips := flips + 1;
fi;
od;
res := res + flips;
all := all + 1;
od;
res/all;
end;
X :=
proc(L)
option remember;
local m, d;
m := nops(L); d := add(p, p in L);
2/d*add(add(L[p]*L[q], q=p+1..m), p=1..m);
end;
An elementary argument is sure to appear now that the answer, which is
very simple, has been posted.
What we have here is essentially the DFA method, a legacy algorithm.
