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If we have a set $\{a,b,c\}$ there are going to be $8$ subsets, namely

one empty subset: $\{\emptyset\}$

$3$ subsets of size one: $\{a\},\{b\},\{c\}$

$3$ subsets of size two: $\{a, b\}, \{a, c\}, \{b, c\}$

$1$ subset of size three: $\{a,b,c\}$

now if one goes to a higher set such as $\{a,b,c,d\}$ and etc, the number of subsets increase and would follow $2^n$ subsets rule. Now I wonder if there is a mathematical pattern to group these subsets by the size of elements exist in each subset as I did in the above example.

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  • $\begingroup$ Did you want the number of subsets of a given length? $\endgroup$ – mvw May 21 '18 at 10:19
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Yeah sure. A number of subsets with cardinality $k$ in a set with cardinality $n$ is given by the binomial coefficient $${n\choose k} = {n!\over k!(n-k)!}$$

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  • $\begingroup$ @Henning Thanks $\endgroup$ – Aqua May 21 '18 at 10:18
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    $\begingroup$ All I would add for my own answer would be some fluff about how the binomial coefficients are important in many areas of mathematics. $\endgroup$ – Henning Makholm May 21 '18 at 10:20
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Pascal's triangle provides an exact answer to this. \begin{array}{rccccccc} n=0: \quad & &&& 1 = \binom{0}{0} &&& \\ n=1: \quad & && 1 = \binom{1}{0} && 1 = \binom{1}{1} && \\ n=2: \quad & & 1 = \binom{2}{0} && 2 = \binom{2}{1} && 1 = \binom{2}{2} & \\ n=3: \quad & 1 = \binom{3}{0} && 3 = \binom{3}{1} && 3 = \binom{3}{2} && 1 = \binom{3}{3} \end{array}

The $n$-th row of the triangle shows how the subsets are split up and the $n$–th row sums to $2^n$ which holds with the number of subsets.

The binomial coefficient $\binom{n}{k}$ gives the number of ways of how to draw $k$ numbers from a set of $n$ elements.

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  • $\begingroup$ You can easily show this is true by considering how you are generating the subsets, e.g for a subset of size 3 you are choosing 3 elements from n etc. $\endgroup$ – Myles Mckay May 21 '18 at 10:15

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