How would i change my answers into a "sensible set theory notation and formulae"? or are my answers fine to this question? 
In a Sixth Form year there are 38 students, each taking at least one of
the subjects of Maths, English, or Science. You know the following
about the number of students enrolled on each of the subjects:

32 students are taking Maths;
26 students are taking English;
31 students are taking Science;
23 students are taking both Maths and English;
27 students are taking both Maths and Science;
24 students are taking both English and Science.

With the information above and
using sensible Set Theory notation and formulae, determine:
1.Whether there are there any students who are taking both Maths and English, but who are not taking Science?
2.How many students are taking both Maths and Science, who are not also taking English?
3.How many students are taking only one subject?
You may find a Venn diagram indicating the sizes of the sets to be
useful.

Here are the answers I got for the following questions:

*

*NONE

*4

*10

Link to my Venn diagram on how I got my answers:

M = Math,
E = English,
and S = Science
So is this how the answer should look? Or is the question asking for a different kind of answer?
 A: Set-theoretic notation is a particular mathematical set of notations devised to state and solve problems like this one. The Encyclopædia Brittanica has a useful article introducing Naïve Set Theory. This version of set theory was developed in the the late 1870s by a group of mathematicians including Georg Cantor, but was later found to have inconsistencies, famously by Bertrand Russell. It has been superseded in modern mathematics by Axiomatic Set Theory, especially by a type called Zermelo-Fraenkel Set Theory, or ZF, but Naïve Set Theory will be more than enough for us.
To restate your problem and solution in set-theoretic terms, we would say something like:

Let $\Omega$ be the set of sixth form students, where $\lvert\Omega\rvert = 38$.

For a set $S$, $\lvert S\rvert$ represents the cardinality or size of the set: the number of elements within the set.

Let $M \subseteq \Omega$ be the subset of sixth-form students taking Maths, $E \subseteq \Omega$ be the subset of students taking English, and $S \subseteq \Omega$ be the subset of students taking science.

$X \subseteq Y$ means that $X$ is a subset of $Y$: every element of $X$ is also an element of $Y$.

We know that $\lvert M\rvert = 32$, $\lvert E\rvert = 26$, and $\lvert S\rvert = 31$. Further, we know that $\lvert M \cap E\rvert = 23$, $\lvert M \cap S\rvert = 27$, and $\lvert E \cap S\rvert = 24$.

$X \cap Y$ is read as "the intersection of $X$ and $Y$", and represents the set containing every element of $X$ that is also an element of $Y$. Thus, $M \cap E$ is the set of all maths students who are also English students.

Determine:

*

*$\lvert M \cap E \setminus S\rvert$

*$\lvert M \cap S \setminus E\rvert$

$X \setminus Y$ is the set-theoretic difference, and is read as "$X$ minus $Y$", and represents the set containing every element of $X$ that is not an element of $Y$. Thus, $M \setminus E$ is the set of all maths students who are not English students.



*$\lvert \Omega \setminus ((M \cap E) \cup (M \cap S) \cup (E \cap S))\rvert$.


$X \cup Y$ is read as "the union of $X$ and $Y$", and represents the set containing every element of $X$ and every element of $Y$. Thus, $M \cup E$ is the set of all maths and all English students. Thus, the complicated expression above is the set of all sixth form students, minus the set of students who are either maths and English students, or maths and science students, or English and science students. This is the set of all students who take only one subject.
A couple of useful set-theoretic formulae are:
$$\lvert A \setminus B \rvert = \lvert A \rvert - \lvert A \cap B \rvert$$
$$\lvert A \cup B \rvert = \lvert A \rvert + \lvert B \rvert - \lvert A \cap B \rvert$$
$$A = (A \setminus B) \cup (A \cap B)$$
We can use the second of these formulae (the principle of inclusion and exclusion) to find that
$$\lvert M \cup E \cup S \rvert = \lvert M \rvert + \lvert E \rvert + \lvert  S \rvert - \lvert M \cap E \rvert - \lvert M \cap S \rvert - \lvert E \cap S \rvert + \lvert M \cap E \cap S \rvert$$
Identifying $\Omega = M \cup E \cup S$, and substituting in known values, we get
$$38 = 32 + 26 + 31 - 23 - 27 - 24 + \lvert M \cap E \cap S \rvert$$
Which grants us $\lvert M \cap E \cap S \rvert = 23$. Thus, 23 students study all three subjects.
From this, we can use arithmetic to get
$$\lvert M \cap E \setminus S\rvert = \lvert M \cap E \rvert - \lvert M \cap E \cap S\rvert = 23 - 23 = 0$$
$$\lvert M \cap S \setminus E\rvert = \lvert M \cap S \rvert - \lvert M \cap S \cap E\rvert = 27 - 23 = 4$$
and
$$\lvert \Omega \setminus ((M \cap E) \cup (M \cap S) \cup (E \cap S))\rvert = \lvert \Omega \rvert - \lvert \Omega \cap ((M \cap E) \cup (M \cap S) \cup (E \cap S))\rvert\\ = 38-\lvert (M \cap E) \cup (M \cap S) \cup (E \cap S)\rvert\\
= 38-\lvert M \cap E \rvert - \lvert M \cap S\rvert - \lvert E \cap S\rvert + 3\lvert M \cap E \cap S \rvert - \lvert M \cap E \cap S \rvert\\
= 38 - 23 - 27 - 24 + 2 \times 23 = 10$$
