ordering balls in slots Three balls are to be placed in 8 slots.
a. If we distinguish among the balls by naming them $ B_1, B_2, B_3 $, in how many different ways can we do this?
I think that this permutation problem, so the solution is
$$
\binom{8}{3}*3!
$$
b. If we do not distinguish among the balls in how many different ways can we do this?
I think that this combination problem, so the solution is
$$
\binom{8}{3}
$$
c. What is the probability that  ball $ B_1, B_2, B_3 $ comes before the other two balls when we examine the slots in a certain order (left to right)?
I couldn't decide how I should consider the solution?
Always one ball comes before other two so i can say that result is equal to 1, on the other hand if question asks the balls porbability separately $B_1$ probability is equal to
$$
\frac{\binom{7}{2}*2!+\binom{6}{2}*2!+\binom{5}{2}*2!+\binom{5}{2}*2!+\binom{4}{2}*2!+\binom{3}{2}*2!+\binom{2}{2}*2!}{\binom{8}{3}*3!}
$$
Is this solutions correct in a and b? And which solution is correct for option c?
Thanks in advance.
 A: 1) Choose 3 slots - this you can do in $\binom{8}{3}$ ways.
Now as the balls are distinguishable you have $3!$ ways to place them in the (already selected) 3 slots. So in total you get $3! \cdot \binom{8}{3}$ ways to do what's wanted.  
2) Choose 3 slots - this you can do in $\binom{8}{3}$ ways.
Now as the balls are not distinguishable you have only 1 way to place them in the (already selected) 3 slots. So you get $1 \cdot \binom{8}{3}$ ways to do this.
3) You have $3! = 6$ ways in which the balls can be ordered. 
B1 B2 B3
B1 B3 B2   
B2 B1 B3
B2 B3 B1   
B3 B1 B2
B3 B2 B1   
Apparently in 2 of these 6 cases B1 comes before the other two Bs
(no matter if we assume left to right or right to left order).    
So the probability is 2/6 = 1/3.
Same is the answer for B2 and B3, the probabilities are still 1/3
(which is not surprising as the problem is symmetric with respect
to B1, B2, B3; I mean none of these Bs is more special than the
other two).  
Btw, 3) implies that the balls are distinguishable.
Otherwise, 3) makes no sense to be asked at all.    
A: While (a) and (b) are correct, I think the answer for (c) is $\frac{1}{3}$. You don't need to consider which slots the balls are in, just think of the order of the three balls from left to right. There are $3! = 6$ permutations, and in $2$ out of the $6$ permutations, $B_i$ comes before the other two.
Another way to think is: As you have said, the probability of any one ball comes before other two is equal to $1$, and by symmetry, there is the same chance for any one of the three balls to come before the other two. Hence the answer is $\frac{1}{3}$.
