# 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 :

1. A) 4/13
2. B) 1/7893600 (that is the Permutation answer)
3. C) 1/120
4. D) 1/65780 (that is the combination answer)
• The critical idea is that it doesn't matter in what order the participants are selected. All you care about is who they are. That is combinations, not permutations. Oct 8, 2012 at 2:24
• that is what I thought ! I just want to confirm it =) Oct 8, 2012 at 2:25

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 $$\frac{5}{26}\cdot\frac{4}{25}\cdot\frac{3}{24}\cdot\frac{2}{23}\cdot\frac{1}{22}.$$