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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)
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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. – Ross Millikan Oct 8 '12 at 2:24
that is what I thought ! I just want to confirm it =) – shnisaka Oct 8 '12 at 2:25
up vote 3 down vote accepted

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

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