Need Explanation with Probability Problem from textbook I am going through probability exercise in a book, and kinda disagree with the reasoning provided by the solution. The answer however is the same. Here is the problem
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. 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?
I know this is asking $P(1 leggings \cap 2 jeans)$, which consists of 


*

*$P(A) = P(pick1=\mathbf{legging} \cap pick2=jeans \cap pick3=jeans)$

*$P(B) = P(pick1=jeans \cap pick2=\mathbf{legging} \cap pick3=jeans)$

*$P(C) = P(pick1=jeans \cap pick2=jeans \cap pick3=\mathbf{legging})$
So, I have to add $P(A) + P(B) + P(C)$. 


*

*$P(A) = P(pick1=\mathbf{legging} \cap pick2=jeans \cap pick3=jeans) = {5\over22} * {6\over23} * {7\over24}$

*$P(B) = P(pick1=jeans \cap pick2=\mathbf{legging} \cap pick3=jeans) = {6\over22} * {5\over23} * {7\over24}$

*$P(C) = P(pick1=jeans \cap pick2=jeans \cap pick3=\mathbf{legging}) = {6\over22} * {7\over23} * {5\over24}$
When looking at the answer, the book is claiming that $3  P(C)$ is the answer. In this case, the multiplication results turns out to be the same. But how do i know instinctively to just pick one from $P(A), P(B) or P(C)$ and multiply it by 3? I cant seem to wrap my head around it. Is it just by looking at the numerators ?(different set of permutation of numerators over the same denominators)
 A: Observe that:
$$ P(A) = P(B) = P(C) $$
This implies that
   $$ 3P(C) = P(A) + P(B) + P(C) $$
A: 7+ 4+ 8= 19 so 24- 19= 5 students ("the rest") are wearing leggings.  The probability of choosing a student wearing leggings at random is  5/24.  Given that there are 23 students left, 7 of them wearing jeans.  The probability the second student chosen is wearing jeans is 7/23.  Given that there are 22 students left, 6 of them wearing jeans.  The probability that the third student chosen is wearing jeans is 6/22= 3/11.  The probability students wearing "leggings, jeans, jeans" are chosen in that order is (5/24)(7/23)(3/11).
The ways of ordering 3 distinct things is 3!= 6 but here the two students wearing jeans are not distinct.  Writing "L" for leggings and "J" for jeans the different orders are "JLL", "LJL", and "LLJ".  There are 3 ($\frac{3!}{2!1!}$) different orders, all with the same probability, (5/24)(7/23)(3/11), so the probability of choosing "one student wearing leggings, two wearing jeans" is 3(5/24)(7/23)(3/11)= 105/2024 which is about 0.052.
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. 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?
