Intuition behind combinatorics problems? I have a hard time understanding when a problem is a combination or permutation. Especially when using the multinomial theorem.
For example: three boxes numbered 1,2 and 3 for k = 1 2 and 3, box k contains k blue marbles and 5-k red marbles. In a 2-step experiment, a box is selected at random and 2 marbles are drawn from it without replacement. If the probability of selecting box k is proportional to k, what is the probability that the 2 marbles drawn have different colors?
As well as: A box contains 35 gems, of which 10 are real diamonds and 25 are fake diamonds. Gems are randomly taken out of the box, one at a time without replacement. What is the probability that exactly 2 fakes are selected before the second real diamond is selected?
I always mess up how to use combinations, I end up getting the number for all possible solutions, which in probability is the denominator. But I can't get the numerator which is always confusing me. I just don't know how to think about these problems in a way to understand them.
 A: The combinatorial element is not strong in these two problems. 
Imagine that the $35$ "gems" have engraved ID numbers on them, to make them distinct.
Way 1: We want the probability that $2$ fakes were drawn before the second real. This could happen with the following patterns of drawing: (i) RFFR, (ii) FRFR, and (iii) FFRR.
We will find the probability of each pattern, and add them together. First we find the probability of the pattern RFFR. The probability the first gem drawn is real is $\frac{10}{35}$. Given that the first was real, the probability the second if fake is $\frac{25}{34}$. Given that the first was real and the second was fake, the probability the  third is fake is $\frac{24}{33}$. And given the first three had pattern RFF, the probability the fourth is real is $\frac{9}{32}$. Thus the probability of (i) is $\frac{10}{35}\cdot\frac{25}{34}\cdot\frac{24}{33}\cdot\frac{9}{32}$.  
If you do the calculations for the probabilities of (ii) and of (iii), you will get products that look different but are equal. Thus our required probability is $3\cdot \frac{10}{35}\cdot\frac{25}{34}\cdot\frac{24}{33}\cdot\frac{9}{32}$.
Way 2: This is a close relative of Way 1, and uses permutations.
There are $(35)(34)(33)(32)$ ways we can draw $4$ gems, where order counts. These ways are equally likely. Now we count the number of favourables, sequences that have $2$ F and $1$ R in the first three, and an R on the fourth. 
The number of patterns of the shape RFFR is $(10)(25)(24)(9)$. Similarly, the number of patterns of the shape FRFR is $(25)(10)(24)(9)$, and the number of patterns of the shape FFRR is $(25)(24)(10)(9)$. The three numbers are all the same, so the number of favourables is $(3)(25)(24)(10)(9)$. Divide by $(35)(34)(33)(32)$ for the probability.

The first problem is somewhat more complicated, at least in wording. 
Box 1 has $1$ blue and $5$ red.
Box 2 has $2$ blue and $10$ red.
Box 3 has $3$ blue and $15$ red. 
The probability of selecting Box $k$ is proportional to $k$. So the probability Box $k$ is selected is $ak$ for some constant $a$. Thus $a+2a+3a=61$. It follows that the probability of selecting from Box 1 is $\frac{1}{6}$, the probability of selecting from Box 2 is $\frac{2}{6}$, and the probability of selecting from Box 3 is $\frac{3}{6}$. 
Let $p_k$ be the probability we select two balls of the same colour given that we draw from Box $k$. Then the required probability is
$$\frac{1}{6}p_1+\frac{2}{6}p_2+\frac{3}{6}p_3.\tag{1}$$
It remains to compute $p_1$ and $p_2$ and $p_3$. 
As a sample, we compute $p_2$. We get balls from the same colour, given that we are sampling from Box 2, in two ways: RR and BB. The probability of RR is $\frac{2}{12}\cdot\frac{1}{11}$. The probability of BB is $\frac{10}{12}\cdot\frac{9}{11}$. Add. 
We leave it to you to compute $p_1$ and $p_3$, and do the computation of (1).
