What is the formula for this combination? Please help me write the formula for all possible combinations given below.
I am not expert in Maths, so please understand.
Elements: a,b,c,d,e
Example of possible combinations:
{a},{b},{c},{d},{e},
{a,b},{a,c},{a,d},{a,e},{b,c},{b,d},{b,e},{c,d},{c,e},{d,e},
{a,b,c},{a,b,d},{a,b,e},{a,c,d},{a,c,e},{a,d,e},
{a,b,c,d},{a,b,c,e},
{a,b,c,d,e}

As you see, no elements are repeating and number of selection is growing from 1 to 5.
I read about Permutations and Combinations basics but couldn't find the right formula.
 A: What you are actually trying to find the number of non-empty subsets of the set $$S=\{1, 2, 3, 4, 5\}$$, and you've listed over half of them.
The number of subsets, including the empty set $\{\;\,\}$, of a set with $5$ elements is given by $2^5 = 32$.  
There are two choices with respect to including an element or not; so there are $$2\times2\times 2 \times 2 \times 2 = 2^5\;\;\text{possible sets}$$ If you want to count only the number of non-empty sets, we then have $2^5 - 1= 32-1 = 31$ subsets.
In general the number of subsets of a set with $n$ elements is given by $2^n$. If we take out the empty set, we count $2^n-1$
Note that in your list of three-element sets, you're missing 
$$\{b,c,d\}, \{b,c,e\}, \{b, d, e\}, \{c, d, e\}$$
And you are also missing 3 four-element sets.
$$\{a, b, d, e\}, \{a, c, d, e\}, \{b, c, d, e\}$$
A: You have $\binom{5}{1}$ ways to choose one element, $\binom{5}{2}$ ways  to choose two elements, $\binom{5}{3}$  ways  to choose three elements, $\binom{5}{4}$  ways  to choose four elements, and, $\binom{5}{5}$  ways to choose five elements.
So, using the binomial expansion, we get
$$\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}=(1+1)^5-1=31.$$
A: First row of combinations:$$\binom51=\frac{5}{1}=5$$
Second row of combinations:$$\binom52=\frac{5\times4}{1\times2}=10$$
Third row of combinations:$$\binom53=\frac{5\times4\times3}{1\times2\times3}=10$$
Fourth row of combinations:$$\binom54=\frac{5\times4\times3\times2}{1\times2\times3\times4}=5$$
Fifth row of combinations:$$\binom55=\frac{5\times4\times3\times2\times1}{1\times2\times3\times4\times5}=1$$
These are the binomial coefficients.  According to the binomial theorem, we have
$$1+\binom51+\binom52+\binom53+\binom54+\binom55=(1+1)^5=32$$
