Question

How many three-letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?

& a)\,990 \\\ & b)2730 \\\ & c)12870 \\\ & d)1560000 \\\ \end{aligned}$$

Answer 1

First determine how many capital English alphabets are symmetric in nature. Now from three cases, in the first case select only one symmetric alphabet and fill the other two places with non-symmetric letters, in the second case select two symmetric alphabet and fill the remaining one place with non-symmetric letter, in the third case select all the three letters as symmetric and fill the three places. Add the number of ways obtained in all the three cases to get the answer.

Complete step by step answer:
Here, we are asked to form three letter computer passwords with no repetition and at least one symmetric letter. We have to determine the possible number of passwords that can be formed. Now let us see how many letters are there which are symmetric. So, there are $11$capital English alphabets which are symmetric in nature, they are: - A, H,I,M,O,T,U,V,W,X,Z. These letters appear to be the same when looked in the mirror and therefore they are called symmetric letters. Here, we need to form the passwords with at least one symmetric letter. So, three cases may arise, let us check them one by one.
1. Case (i): - only one symmetric letter: -
Here we have to select one symmetric letter. So, we have,
Number of ways to select one symmetric letter from the $11$ symmetric letters present = $^{11}{{C}_{1}}=11$
Since, we have to fill three places, so the other two places must be filled with two of the remaining $15$ asymmetric letters.
Number of ways to select two asymmetric letters from $15$ such letters= $^{15}{{C}_{2}}=\dfrac{15\times 14}{2!}=105$
\begin{matrix} \\_ & \\_ & \\_ \\\ \end{matrix}
Now, the selected letters can be arranged at three places in $3!$ ways. So, the number of words formed in the first case will be $11\times 105\times 3!=6930$.
2. case (ii) two symmetric letters.
Here, we have to select two symmetric letters and one asymmetric letter so we have,
Number of ways to select two symmetric letters from $11$ symmetric letters present = $^{11}{{C}_{2}}=55$.
Number of ways to select one asymmetric letter from $15$ such letters = $^{15}{{C}_{1}}=15$.
\begin{matrix} \\_ & \\_ & \\_ \\\ \end{matrix}
Now, the selected letters can be arranged at three places in $3!$ ways. So, the total number of words formed in the second case will be $55\times 15\times 3!=4950$
3. case (iii) three symmetric letters:
Here, we have to fill all the three places with symmetric letters, so we have to select three symmetric letters. So, we have,
Number of ways to select three symmetric letters from $11$ symmetric letters present = $^{15}{{C}_{3}}=65$
\begin{matrix} \\_ & \\_ & \\_ \\\ \end{matrix}
Now, the selected letters can be arranged at three places in $3!$ ways. So, the total number of words formed in the third case will be $165\times 3!=990$
Therefore, the number of possible three- letter computer passwords that can be formed with the conditions provided will be given as: $6930+4950+990=12870$

So, the correct answer is “Option c”.

Note: One must note that we have to form only three letter passwords so no more cases can be formed according to the conditions provided. we cannot form a password with three asymmetric letters because it is given in the question that the password must contain at least one symmetric letter. You have to remember the number of symmetric letters to solve the question. Remember the formulas of combination and arrangements.