# How many subsets with an odd number of elements does a set with 10 elements have?

This question aims to find out how many combinations of a set with ten elements could be made. We need to build our understanding of a basic concept of combination for that purpose.

Moreover, this question is based on the concepts of statistics. A set is a well-defined collection of different things which may include books, pens, students, etc. In combination, without considering the order of a set, all of the specific parts in a set are selected.

A subset has $n$ elements of a set in which there are $r$ – combinations of these $n$ elements. Mathematically, the combination of $n$ elements can be found as follows.

$C( n, r ) = \dfrac {n!}{r! (n – r)! } \text{ with }n \ne n . (n – 1). (n – 2). … .2 . 1$

We are only interested to find the odd number subsets that a set have with 10 elements. Therefore:
$n = 10$

$r = 1, 3, 5, 7, \text{ or, } 9$

and the total number of subsets are:

$\text{Number of subsets} = \sum_{r\in{{1, 3, 5, 7, 9 } }^{} } C(10, r)$

$= C(10, 1) + C(10, 3) + C(10, 5) + C(10, 7) + C(10, 9)$

$= \dfrac{10!}{1! (10 – 1)!} + \dfrac{10!}{3! (10 – 3)!} + \dfrac{10!}{ 5! (10 – 5)!} + \dfrac{10! }{ 7! (10 – 7)!} + \dfrac{10!}{9! (10 – 9) !}$

$= \dfrac{10!}{1! \times 9!} + \dfrac{10!}{3! \times 7!} + \dfrac{10!}{5! \times 5! } + \dfrac{ 10! }{7! \times 3!} + \dfrac{10!}{9! \times 1!}$

Since:

$n! = (n – 1) \times (n – 2) \times … 3. 2. 1$

$= 10 + 120 + 252 + 120 + 10$

$= 512$

## Alternative Solution

A set having $n$ elements contains a total $2^n$ number of subsets. In these subsets, half of the numbers have odd cardinality, and half have positive cardinality.

Therefore, an alternative solution to find the number of subset in a set with an odd number of elements are:

$\text{Number of subsets} = \dfrac{2^n}{2}$

$= 2^{n – 1}$

$= 2^9$

$= 512$

## Numerical Results

The number of subsets with an odd number of elements does a set with 10 elements have:

$\text{Number of subsets} = 512$

## Example

Find the subsets of the first eight prime numbers.

Solution:

The set of first 8 prime numbers are as follows:

$p = {1, 3, 5, 7, 11, 13, 17, 19}$

As the total number of subsets is $2^n$, wherein our set has $n = 8$ elements.

Therefore, the number of subset of a set containing first eight prime numbers as elements are:

$\text{Number of subsets} = 2^8$

$= 256$

Images/ Mathematical drawings are created with Geogebra.