Probability of selection containing certain options Question 3.27 from the book Openintro statistics confuses me. The question states:

In a classroom with 24 students, 7 students are wearing jeans, 4 are wearing shorts, 8 are wearing skirts, and the rest are wearing leggings(5). If we randomly select 3 students without replacement, what is the probability that one of the selected students is wearing leggings and the other two are wearing jeans? Note that these are mutually exclusive clothing options.

I tought the answer would be $\frac{5}{24} \times \frac{7}{23} \times \frac{6}{22} = 0.0173$
However, the answer states:

$\frac{5}{24} \times \frac{7}{23} \times \frac{6}{22} = 0.0173$. However, the person with leggings could have come 2nd or 3rd, and these each have this same probability, so $ 3 \times 0.0173 = 0.0519$

This confuses me greatly. How would you for example compare this to the probabilty of drawing 3 hearts from a shuffled deck of cards, assuming no replacement. This would be $\frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} = 0.0129$ But here we don't multiply by 3, why? What about the probability of drawing 2 hearts and 1 diamond?
 A: The answer is a probability with the number of 'favorable' outcomes
in the numerator and the number of 'all possible' outcomes in the denominator. In many cases, you have your choice whether 'outcomes'
are ordered (order matters) or unordered (doesn't matter). However, once
you make your choice, you need to be consistent, using either ordered
or unordered outcomes both in numerator and denominator.
A method with unordered outcomes is as follows:
$$P(\mathrm{2\, Jeans\, and\, 1\, Leggings}) =
\frac{{7\choose 2}{5\choose 1}{12\choose 0}}{{24\choose 3}}
=  0.05187747.$$
Computation in R:
choose(7,2)*choose(5,1)/choose(24,3)
[1] 0.05187747

Note: By general agreement, for most card games, outcomes are
regarded is unordered. (The order in which cards are dealt has no effect on the value of the hand.)  So a straightforward combinatorial solution
uses unordered outcomes in numerator and denominator.
Thus, getting three hearts (taken as indistinguishable) when dealing three cards from a well-shuffled deck is ${13\choose 3}{39\choose 0}/{52\choose 3}= 0.01294118.$
choose(13,3)/choose(52,3)
[1] 0.01294118

