Please check my attempt for correctness:
choosing 3 out of 50 children can be done in 50C3 ways.
And to satisfy the condition, i can make a selection of 1 from 13, 1 from 25 and 1 from 12 students.
so i have 12*13*25/(50C3).
There are $1000$ candies, of which $400$ are Skittles (S) and $200$ Almond Jpys (AJ). We can analyze the problem in two ways: (i) Imagine the picks are consecutive or (ii) Imagine that the householder grabs simultaneously two pieces of candy.
The probability the first pick is skittles is $\dfrac{400}{1000}$. If the first pick was S, the probability the next is AJ is $\dfrac{200}{999}$. So the probability of S then AJ is $\dfrac{400}{1000}\cdot \dfrac{200}{999}$.
Similarly, the probability of AJ then S is $\dfrac{200}{1000}\cdot\dfrac{400}{999}$, the same. So our desired probability is
$$2\dfrac{400}{1000}\cdot \dfrac{200}{999}.$$
We could also count. Imagine the candies all have ID numbers. There are $(1000)(999)$ equally likely ways of picking two candies, counting order. And there are $(400)(200)$ ways of picking S, then AJ. There are $(200)(400)$ ways to pick AJ, then $S$. So our probability is $\dfrac{(400)(200)+(200)(400)}{(1000)(999)}.$
For way (ii), there are $\dbinom{1000}{2}$ ways to pick a pair of candies. And there are $(400)(200)$ ways to end up with one S and one AJ. Divide.
For the second problem, again we could count order or not, and we could work directly with probabilities, or count.
We will do it by looking at the $(1000)(999)$ ways of picking two candies, in order.
There are $(100)(99)$ ways to pick two Kit-Kats, $(200)(199)$ to pick two Almond Joys, and so on. So our probability is
$$\dfrac{(100)(99)+(200)(199)+(300)(299)+(400)(399)}{(1000)(999)}.$$
Or else one can say that there are $\dbinom{1000}{2}$ ways to pick two candies, order of picking not counting. There are $\dbinom{100}{2}$ ways that these two are Kit-Kats, $\dbinom{200}{2}$ ways they are Almond Joys, and so on. When we calculate this way, we end up he same answer, though it looks a little different.
Best Answer
Since all the answer asks for is plain M&Ms vs not plain M&Ms, we can simplify the information to 3 packages of plain and 5 packages of not plain.
Now the number of ways we can choose 1 plain package from 3 possibilities is $3\choose1$ and the number of ways to choose 1 not plain package from 5 possibilities is $5\choose1$.
From this, we find that the number of ways to choose one plain package and one not plain package is $${3\choose1} {5\choose1} = 15$$
Now, the total number of ways to choose 2 packages from 8 possibilities is $${8\choose2} = 28$$ so the probability of picking the desired set of packages is $\frac{15}{28}$.