Statistic (Combination and Permutation) I have this problem which I could not figure out if I should do it by using Combination or Permutation

The Organizer of a television show must select 5 people to participate
  in the show. The participants will be selected from a list of 26
  people who have written in the show. If the participants are selected
  randomly, what is the probability that the 5 youngest people will be
  selected

and I have those choices : 


*

*A) 4/13

*B) 1/7893600 (that is the Permutation answer)

*C) 1/120

*D) 1/65780 (that is the combination answer)

 A: There are $\binom{26}{5}$ ways to select $5$ people. All these ways are equally likely. Exactly one of these ways results in choosing the $5$ youngest. So the required probability is 
$$\frac{1}{\binom{26}{5}}.$$
We can also produce the correct answer using permutations. Imagine selecting the people in order. There are $(26)(25)(24)(23)(22)$ ways to do this, all equally likely.
There are $(5)(4)(3)(2)(1)$ ways to select the five youngest people, in some order. So our probability is
$$\frac{(5)(4)(3)(2)(1)}{(26)(25)(24)(23)(22)}.$$
Remark: The point is that the numerator and the denominator must each count the same type of thing: selections without order, or selections with order. In the second expression, if we use numerator $1$ instead, as in Choice 2., then we are mixing types, counting with order in the denominator but not in the numerator. 
We can get to the answer in a somewhat different way. The probability that the first person chosen is among the $5$ youngest is $\frac{5}{26}$. Given that such a person is selected, the probability the next person chosen is among the (original) $5$ youngest is $\frac{4}{25}$. And so on. So our probability is
$$\frac{5}{26}\cdot\frac{4}{25}\cdot\frac{3}{24}\cdot\frac{2}{23}\cdot\frac{1}{22}.$$
