A problem regarding probability exercises. I have run into a thinking error while trying to solve the following two exercises, here they are:
1) There are 30 fish in the lake. 5 of them have been taken out and marked and then put back into the lake. Later 7 fish have been picked from the lake. What is the probability that 2 out of 7 picked fish were marked?
2) A group of tourists which consists of 15 boys and 5 girls are to participate in a tournament. They need to pick 4 people out of all. What is the probability that their team will consist out of 2 boys and 2 girls?
I have no problems regarding combinatoric calculations, however, there is a logical conundrum that I fail to overcome. 
1) In my math book, it is explained that in the first problem you have to find the total number of possible outcomes using formula A(30,7)= 30!/7! which then says that the order in which we take out our fish is important (A, B, C) and (C, B, A) group of fishes should be regarded as different and calculated as three separate variants. The same goes for the number of favorable outcomes which is found with formula A(5,2)C(25,5)=(5!/3!) (25/20!). The results of our probability is 253/20358.
2) In this exercise, it is explained that when trying to find the number of all the possible outcomes we use formula C(20,4)=20!/(4!*16!) and the number of favorable outcomes if C(15,2)*C(5,2)=(15!/(2!*13!)) * (5!/(2!*3!)). When you divide two the answers is 70/323.
My question is: What makes these two problems so different, that in the first one you have to count all the possible outcomes concerning the order of a group of elements (which means that group (A, B) A is fish number 1 and B is fish number 2 and (B, A) are two different possible outcomes) and in the second exercise there is no difference between a group (A, B) and (B, A) A is boy number 1 and B is boy number 2 and they are counted as the same outcome?
 A: When doing probability, the important thing is to consider order (or not) in both numerator (favorable outcomes) and denominator (total outcomes).
Regarding the first problem. If order is important, then total outcomes are 
$$A_7^{30}=\frac{30!}{23!}$$
and favorable outcomes are
$${7\choose2}A_2^5A_5^{25}=\frac{7!}{2!5!}\frac{5!}{3!}\frac{25!}{20!}$$
The ${7\choose2}$ decide where the marked fish are taken. The probability is $0.2609\ldots$
If order is not important, then total outcomes are 
$${30\choose7}=\frac{30!}{7!23!}$$
and favorable outcomes are
$${5\choose2}{25\choose5}=\frac{5!}{2!3!}\frac{25!}{5!20!}$$
The probability is, again, $0.2609\ldots$
This probability may seem high, but it was expected. This is used in biology to evalutate a fish population in an area. The idea is that the ratio of marked/unmarked fish taken the second time should be the same as the ratio of taken/population the first time. With this example, $5$ marked fish in a population on $30$, should give
$$\frac M7=\frac5{30}\implies M=1.166\ldots$$
With $7$ fish the second time, we expect to have a bit more than $1$ marked. So $2$ is highly probable.
