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Formula

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Results

Total number of subsets
32
= 2^5 (includes the empty set and the full set)
Proper / non-empty subsets (2ⁿ − 1) 31
Subsets of size k = 2 — C(n,k) 10

What is the Subset Calculator?

A subset is any selection of elements taken from a larger set, including the empty set and the set itself. This calculator answers two common combinatorics questions: how many subsets does a set of n elements have in total, and how many of those subsets contain exactly k elements. It is useful for students learning set theory, probability, and discrete mathematics, as well as anyone counting possible combinations.

A three-element set with all eight of its subsets shown as small grouped circles
A 3-element set has \(2^3 = 8\) subsets, including the empty set and the full set.

How to use it

Enter the size of your set as n (the number of distinct elements). The calculator immediately returns the total number of subsets, equal to \(2^n\), and the number of proper/non-empty subsets, \(2^n - 1\). Optionally enter a subset size k to also get the number of subsets containing exactly k elements, computed with the binomial coefficient \(C(n, k)\). Leave k blank if you only need the total.

The formula explained

Each element of the set can independently be included or excluded from a subset — that's 2 choices per element. With n independent elements the total count is $$2 \times 2 \times \dots \times 2 = 2^n$$ To count subsets of a fixed size k, we use the combination formula $$C(n, k) = \frac{n!}{k!\,(n - k)!}$$ which selects k elements without regard to order. Summing \(C(n, k)\) over all k from 0 to n recovers \(2^n\).

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Diagram showing total subsets equals 2 to the n and a single k-sized subset chosen from n elements
Total subsets grow as \(2^n\), while \(C(n,k)\) counts subsets of one fixed size k.

Worked example

Suppose n = 5. The total number of subsets is \(2^5 = 32\), and the number of non-empty subsets is 31. The number of 2-element subsets is $$C(5, 2) = \frac{5!}{2! \cdot 3!} = \frac{120}{2 \cdot 6} = 10$$ So from a 5-element set there are exactly 10 possible pairs.

FAQ

Does the total include the empty set? Yes. The \(2^n\) count includes both the empty set and the full set. Subtract 1 for non-empty subsets, or 2 for proper non-empty subsets.

What if k is larger than n? There are no such subsets, so \(C(n, k) = 0\) whenever \(k > n\) or \(k < 0\).

Why is the max around 170? \(2^n\) and factorials grow extremely fast; beyond roughly n = 170 the values exceed the range that standard floating-point numbers can represent.

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