Question

Four alphabets E, K, S and V one in each, were purchased from a plastic warehouse. How many ordered pairs of alphabets, to be used as initials can be formed from them?

Answer 1

Hint: Here, we need to find ordered pairs so basically choosing two alphabets at a time out of 4.

Complete step-by-step answer:
Given,
Four alphabets E, K, S and V one in each, were purchased from a plastic warehouse.
We know that, the number of arrangements of n things taken r at a time is given by
${}^n{P_r} = \dfrac{n!}{(n-r)!}$
The required number of ordered pairs of alphabets, to be used as initials, can be formed as (or the number of arrangements of 4 things taken 2 at a time)
${}^4{P_2} = \dfrac{4!}{(4-2)!} = \dfrac{{4!}}{{2!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 12$

So this is your desired answer.

Note: Ordered pairs is a group of two alphabets. Here we use permutations because (E, K) and (K, E) are considered as different ordered pairs.