Calculating chance of all heads in 5 toin cosses without multiplication? 5 toin cosses, chance of getting all heads? I know you can easily use multiplication since they are independent events (each toss is), but at this point in the textbook we have not covered it. 
I am only allowed to use additive property (probability of union of disjoint events is the sum of the probability of each event)
and inclusion-exclusion principle $$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$ as well as set operations like union, complement etc.
I know one possible way, which is say each toin coss is event A,B,C,D,E, then
have a complicated expression using inclusion expression principle with many terms?
 A: There are ${5\choose 1}=5$ outcomes with $1$ head. Similarly, there are ${5\choose 2},{5\choose 3},{5\choose 4},{5\choose 5}$ outcomes with $2,3,4,5$ heads, respectively. Hence, the probability of all $5$ coins being head is:
$$\frac{{5\choose 5}}{{5\choose 1}+{5\choose 2}+{5\choose 3}+{5\choose 4}+{5\choose 5}}=\frac{1}{2^5}.$$ 
Note: "toin cosses" must be "coin tosses". Fun facts: "tiyin" is an Uzbek coin and "tashla" is "to toss" in Uzbek.
A: I'm going to demonstrate how to solve this using a careful application of the fundamentals from combinatorial analysis and probability theory.
Given some arbitrary probability experiment with a finite sample space, $S$, and some event $E$ such that $E \subseteq S$ ($E$ is a subset of $S$). It can be shown that $$P(E)= \frac{|E|}{|S|}$$ where $|E|=$the number of outcomes in $E$ and $|S|$ is the number of outcomes in $S$, but ONLY IF each possible outcome in $S$ is equally likely to occur. In the situation you are describing, each possible sequence of results from $5$ coin flips is equally likely to occur, so we are justified in using the above formula. We just need some way of counting the number of outcomes in $S$ and $E$. This is where a good foundation in combinatorial analysis becomes invaluable. First, we count the number of outcomes in $S$.

Counting Outcomes in $S$:
Any time you are counting the number of outcomes in some situation, it is helpful to "break up" the situation into multiple other situations for which the number of outcomes can be easily counted. So, instead of counting the number of outcomes resulting from $5$ coin flips, lets count the number of outcomes resulting from each flip of the coin. When we flip the coin once, there are $2$ outcomes: H or T. When we flip the coin a second time, there are $2$ outcomes: H or T. When flip it a third time, there are $2$ outcomes$...$ for each of the $5$ times we flip the coin, there are $2$ outcomes. 
Now, how do we combine this information to count the number of outcomes after $5$ flips of the coin? Well, we have to determine whether the situation we're dealing with satisfies the generalized principle of counting. To put it simply, we have to determine whether every outcome in the second flip is possible for every outcome in the first flip. Then, we must determine whether every outcome in the third flip is possible for every outcome in the first two flips. Then we must determine whether every outcome in the fourth flip is possible for every outcome in the first three flips. And finally, we must determine whether every outcome in the fifth flip is possible for every outcome in the first four flips. If this is the case, then the generalized principle of counting tells us that we can count the number outcomes resulting from the combination of all $5$ flips by multiplying the number of outcomes resulting from each of the individual flips. Hence,
$$|S|=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2=2^5=32$$

Counting Outcomes in $E$:
Our event $E$ is a subset of $S$ consisting of all outcomes for which all coin flips are H. We can apply the same logic as above, but this time applying our criteria that all coin flips must be H. 
When we flip the coin once, there is $1$ outcome for which the result is H. When we flip the coin a second time, again there is $1$ outcome for which the result is H. When flip it a third time$...$ for each of the $5$ times we flip the coin, there is $1$ outcome for which the result is H. 
Now we must evaluate whether the individual flips we're dealing with satisfy the generalized principle of counting. Certainly, any of the outcomes from one arbitrary flip are possible for each of the outcomes resulting from all the flips before. So, 
$$|E|=1 \cdot 1 \cdot 1 \cdot 1 \cdot 1=1^5=1$$

Now that we have solved for $|S|$ and $|E|$, we solve by substitution: 
$$P(E)= \frac{|E|}{|S|}=\frac{1}{32}=0.03125$$
Hence, the probability of flipping a coin $5$ times and getting nothing but H is $0.03125$
