Each group gets $6$ out of $18$ questions ($3$ easy, $15$ hard), probability all groups have one easy question 
In a course, students form three groups. Each group is assigned $6$ questions, chosen at random, without replacement,  from a set of $18$ questions. From these $18$ questions, $3$ are easy and $15$ are hard. What is the probability all groups have one easy question?

Attempt. Let $E_i$ be the event group i has at least one easy question. The desired event is $E_1E_2E_3$ and using complements and inclusion/exclusion:
$$\mathbb{P}(E_1E_2E_3)=1-\mathbb{P}(E_1'\cup E_2'\cup E_3')=1-
\big[3\mathbb{P}(E_1')-3\mathbb{P}(E_1'E_2')+\mathbb{P}(E_1'E_2'E_3')\big]$$
(since $\mathbb{P}(E_i')=\mathbb{P}(E_j')$ and $\mathbb{P}(E_i'E_j')=\mathbb{P}(E_k'E_m')$). Also:
$$\mathbb{P}(E_1')=\frac{\binom{3}{0}\binom{15}{6}}{\binom{18}{6}},$$
$$\mathbb{P}(E_1'E_2')=\frac{\binom{3}{0}\binom{15}{12}}{\binom{18}{12}},$$
$$\mathbb{P}(E_1'E_2'E_3')=0.$$
Do you see any flaw in the above argument?
Thanks in advance.
 A: Your calculation is correct.  We can confirm this by calculating the probability directly.
There are
$$\binom{18}{6}\binom{12}{6}\binom{6}{6}$$
ways to distribute the cards to three labeled groups.
If each group receives one easy question, then there are $3!$ ways to distribute the easy questions and $\binom{15}{5}\binom{10}{5}\binom{5}{5}$ ways to distribute the hard questions, so there are
$$3!\binom{15}{5}\binom{10}{5}\binom{5}{5}$$
ways to distribute the questions to three labeled groups so that each group receives one easy question.
Hence, the probability that each group receives one easy question is
$$\frac{3!\dbinom{15}{5}\dbinom{10}{5}\dbinom{5}{5}}{\dbinom{18}{6}\dbinom{12}{6}\dbinom{6}{6}}$$
which agrees with your answer.
A: There's another way to approach this.  Instead of assigning questions to groups, assign groups to questions.  In other words, look at which group each question is assigned to.
Assign the first easy question to a group arbitrarily.  For this to be a "good" assignment, there is a probability of $\frac{12}{17}$ that the second easy question will be assigned to a different group.  That's because there remain $17$ vacant spaces of which $12$ are good.  If that's the case, there is a $\frac{6}{16}= \frac 38$ probability that the third easy question will be assigned to the third group because now there are $16$ remaining vacant spaces of which $6$ are good.  Thus, the probability of a "good" assignment is $\frac{36}{136}=\frac{9}{34}$.
