Question

A special type of password consists of four different letters of the alphabet, where each letter is used only once. How many different possible passwords are there?
A. \[{{4}^{26}}\]
B. \[456976\]
C. \[14950\]
D. \[358800\]

Answer 1

Hint: Password consists of four letters of the alphabet and we know that there are a total of 26 alphabets in the English language. To get the total possible cases, we have to use the concept of permutation. The formula we will use is \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\] , here we have n = 26 and r = 4.

Complete step-by-step solution:
Let us first write down the given data as below,
Number of letters contained by a special type of password = 4
No letter is repeated twice in the password.
Now, we know that we have a total of 26 alphabets. So, this is a problem of permutation. To understand the problem let’s take an example of how these 4 letters special type of password can be formed of alphabets.
As there are total 26 alphabets which are given as follows,
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
To form a password of 4 different letters, we take any 4 letters of the above alphabets with the condition that no letter is used more than once.
For example, we can take four letters E F G H of 26 alphabets and these can be arranged in \[{}^{4}{{P}_{4}}\] ways.
So, we can take any 4 letters of 26 alphabets to form the special type of password and the total number of such passwords is calculated by using permutation.
Formula for permutation \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]
Total number of passwords \[={}^{26}{{P}_{4}}\]
\[=\dfrac{26!}{\left( 26-4 \right)!}\]
\[\begin{align}
  & =\dfrac{26!}{22!} \\
 & =\dfrac{26\times 25\times 24\times 23\times 22!}{22!} \\
 & =26\times 25\times 24\times 23 \\
\end{align}\]
Therefore, we get the total number of passwords as \[26\times 25\times 24\times 23=358800\]
Hence, the required number of total special types of passwords is 358800.
Option D is correct.

Note: In the problem related to permutation and combination just be clear about when to use permutation and when to use combination and also remember the formula for them as they are different things. When students write the formula for permutation, they tend to make the mistake in formula as \[{}^{n}{{P}_{r}}=\dfrac{r!}{\left( n-r \right)!}\] instead of \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\] , this is a very silly mistake which can make the whole solution incorrect, so avoid this. We can also think of an alternative way. We can choose 4 different letters out of 26 alphabets in ${}^{26}{{C}_{4}}$ ways. Now we have to arrange the four letters in all possible ways, so we can do that by writing it as $4!\times {}^{26}{{C}_{4}}$ and this will give us
$\begin{align}
  & 4!\times \dfrac{26!}{4!\times 22!} \\
 & \Rightarrow \dfrac{26!}{22!} \\
 & \Rightarrow \dfrac{26\times 25\times 24\times 23\times 22!}{22!} \\
 & \Rightarrow 26\times 25\times 24\times 23 \\
 & \Rightarrow 358800 \\
\end{align}$