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?

Hint: We start solving the problem by finding the total number of ways to choose the two alphabets from the given four in order to make an ordered pair using the fact that the total no. of ways to choose ‘r’ objects out of ‘n’ objects is ${}^{n}{{C}_{r}}$ ways. We then find the total number of ways that the chosen alphabets can be arranged and we multiply this result with the total no. of ways of choosing 2 alphabets to get the required answer.

Complete step-by-step solution
According to the problem, we have four alphabets E, K, S, and V and we need to find the total number of ordered pairs of alphabets formed using these four alphabets which were to be used as initials.
We know that the ordered pair consists of 2 elements. So, we need to choose two alphabets from the four alphabets in order to use them as initials.
We know that the total no. of ways to choose ‘r’ objects out of ‘n’ objects is ${}^{n}{{C}_{r}}$ ways.
So, we can choose 2 alphabets out of 4 alphabets in ${}^{4}{{C}_{2}}$ ways.
We know that the chosen two alphabets can be arranged in two ways. i.e., if we choose alphabets E, K for initials, then we get the initials as $\left( E,K \right)$ and $\left( K,E \right)$.
So, we get the total number of ordered pairs of alphabets to use as initials in ${}^{4}{{C}_{2}}\times 2$ ways.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
So, we get ${}^{4}{{C}_{2}}\times 2=\left( \dfrac{4!}{2!2!} \right)\times 2$.
$\Rightarrow {}^{4}{{C}_{2}}\times 2=\left( \dfrac{4\times 3}{2\times 1} \right)\times 2$.
$\Rightarrow {}^{4}{{C}_{2}}\times 2=4\times 3$.
$\Rightarrow {}^{4}{{C}_{2}}\times 2=12$.
So, we have got that we can make 12 initials by using the given four alphabets.
$\therefore$ The total number of initials that can be formed using the given four alphabets is 12.

Note: We can also solve this problem by assuming that we have to make a word containing 2 alphabets from the 4 alphabets given in the problem (word need not be meaningful in this case). We should not stop solving the problem after finding the total no. of ways of choosing the alphabets as the arrangement is also important in finding the initials. Similarly, we can also expect problems to solve the problem by giving more than 4 alphabets and finding the initials.