Feller - Question on two throws of three dice I am stuck at the following question.

What is the probability that two throws with three dice each will show the same configuration if (a) the dice are distinguishable (b) they are not.

Solution.
(a) $P(\text{Two throws of three dice s.t. each show the same configuration})=6^{3}/6^{6}=1/216$.
(b)Firstly from the second set of throws, the $3$ dies with the same face value as the first ones can be chosen in $3!=6$ ways.
Place $r=3$ dies in $n=6$ cells, such that order does not matter with repetition, can be done in ${{3+6-1}\choose{3}}={8\choose 3}=56$
${\displaystyle P(\text{Two throws of three indistinguishable dice show the same configuration})=\frac{56\times6}{6^{6}}}$
However, my answer to the second part of the problem is incorrect. Could someone help me think correctly about the problem.
 A: (b)
Three distinct faces :
$$P(\text{getting the same configuration}) = \frac{(6\cdot5\cdot4)\times(3\cdot2\cdot1)}{6^6}$$
One repetition:
$$P(\text{getting the same configuration}) = \frac{{3\choose2}(6\cdot5)\times{3\choose2}}{6^6}$$
Two repetitions:
$$P(\text{getting the same configuration}) = \frac{6}{6^6}$$
Thus, the required probability is,
$$P(\text{getting the same configuration}) = \frac{996}{6^6}$$
A: Simulation: Two successive throws of three fair dice with faces 1 through 6. Probability of same three faces showing: (a) if dice are indistinguishable,
(b) distinguishable.
In the distinguishable case, it may help to imagine the dice are colored
red, green, and blue, respectively. The probability is pretty clearly
$1/6^3 =  0.00462963.$ In simulating for the indistinguishable dice, we sort the 
outcomes before comparison because color does not matter.
m = 10^6;  same.i = same.d = numeric(m)
for (i in 1:m) {
  roll.1 = sample(1:6, 3, repl=T)
  roll.2 = sample(1:6, 3, repl=T)
  same.d[i] = sum(roll.1==roll.2)
  same.i[i] = sum(sort(roll.1)==sort(roll.2)) }
mean(same.d==3);  mean(same.i==3)
## 0.004632
## 0.021579

With a million iterations, probabilities should be accurate to 
three places (a little more for very small probabilities), so
the first simulated answer is consistent with $1/63.$ And, the
second answer is consistent with with @Quasar's $996/6^6 = 0.02134774.$ (+1)
For a little more detail, my same.d and same.i are numbers
of faces that agree for distinguishable and indistinguishable dice, respectively.
here are the histograms of their respective distributions.
(The probabilities mainly at issue here are the ones for 3 matches.)

