Probability and combinatorics problems - picking balls and choosing postcards I am sorry for bothering you with such a trivial and easy question (in comparison to the others asked here) but I have no idea where else I could ask. These are two problems I have to solve and I just need you to check if my solution is correct.
First: I have 10 red balls and 3 blue balls. What is the probability of picking 1 red ball and 3 blue balls?
My solution: 10/13 . 3/12 . 2/11 . 1/10 = 1/286
Or should it be 10/13 . 3/12 . 3/11 . 3/10 = 9/572?
Second: They have 16 different postcards in the shop. How many possibilities do we have if we want to choose 6 different ones?
My solution: 16 . 15 . 14 . 13 . 12 . 11 = 5765760
Thanks in advance for checking my solutions
 A: Your method for the first problem is the probability of selecting a red, followed by $3$ blues, sequentially. We have to take into account all the different ways to arrange $3$ blues and $1$ red. Note that this is a hypergeometric random variable, giving probability
$$\frac{{10 \choose 1}{3 \choose 3}}{{13 \choose 4}}\approx 0.014$$
For the second problem, your solution would be correct if ordering mattered, but it does not. Getting $123456$ is considered the same as $654321$. We have
$${16\choose6}=8008$$
A: Both your solutions are incorrect.  Let me start with the second.
You are overlooking the fact that the order in which you choose the postcards is irrelevant.  Picking ABCDEF is the same as picking CDEBFA.  Since there are $6!=720$ ways to arrange any set of six postcards in order, you have counted each set $720$ times.  Divide your solution by $720$ to get the right answer.
In the first problem, you are imposing an order, assuming that we pick the red ball first, then the blues balls.  Perhaps you interpret the question to mean that, but I'm sure it's not what is intended.  There are only $10$
ways to pick the balls, because you must pick all $3$ blue balls, but you can pick any one of the $10$ red ones.  You must divide this by the number of ways to pick $4$ balls from $13$.  This is just like the postcard problem:$$\frac{13\cdot 12\cdot 11\cdot 10}{4\cdot 3\cdot 2 \cdot 1} $$ so we get $$\frac{10\cdot 24}{13\cdot 12\cdot 11\cdot 10}$$
