How many ways can two letters be selected from five distinct letters when order does not matter?

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combinatorics Figure 1: Ferrers' partitioning diagram for 14.

In combinatorics, when selecting two letters from five distinct letters where the order of selection does not matter, the process is defined as a combination.

Because the order is not a factor, subsets such as AB and BA are considered identical. To calculate the number of ways to choose $k$ objects from a set of $n$ objects, we use the formula for combinations, denoted as $_{n} C_{k}$ :

$_{n} C_{k} = \frac{n!}{k! (n - k)!}$

For this specific case, where $n = 5$ and $k = 2$ , the calculation is:

$_{5} C_{2} = \frac{5!}{2! (5 - 2)!} = \frac{120}{2 \times 6} = 10$

There are 10 unique ways to select two letters from five distinct letters: AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE.1