Four letters E, K, S and V, one in each, were purchased from a plastic warehouse.

Here, we need to find out the number of pairs of the letters that can be formed with the 4 letters.

Required number of ordered pairs = Number of arrangements of  four letters, taken two at a time = 4P2

$=\frac{4 !}{(4-2) !}$

$=\frac{4 !}{2 !}$

$=\frac{4 \times 3 \times 2 !}{2 !}$

$=4 \times 3$

$=12$