Question

Capital English alphabet has 11 symmetric letters that appear the same when located in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric three letter passwords can be formed using these letters?

Answer 1

Hint: Question is based on the concept of combination topic. As in the question, a possible number of combinations are asked such that 3 letter passwords can be formed out of 11 symmetric letters which are also symmetric. So according to concept of combination, if we want to find out the possible number of combination without any repetition than we can go with formula of$\dfrac{n!}{r!\left( n-r \right)!}$ (represented as $^{n}{{C}_{r}}$)in which n is the total number of elements, and r is the combination in a letter, as in our case it is 3 letters.

Complete step by step answer:
So moving ahead with the question, we had 11 symmetric elements. And we had to find out the three letter password that should also be symmetric. So we can directly say that there can be one type of arrangement in a password in which all three letters (which are symmetric) are the same. That can be AAA, HHH, III,.......... YYY. As these passwords will always be symmetric. So now there are 11 password combinations we got.
Other than this if we have two symmetric alphabets at the end and one asymmetric alphabet in the middle which is different to other two alphabets at the end, this combination will also be symmetric. For example; HAH, IAI, AHA, YWY,.......... AYA. So we can write it as a combination of, at the ends of 11 symmetric alphabet 1 alphabet and from the remaining 10 remaining symmetric alphabet we want any 1. So relating it with the concept of combination we can write it as$^{11}{{C}_{1}}{{\times }^{10}}{{C}_{1}}$. So we can write it as;
\[\begin{align}
  & =\left( \dfrac{11!}{1!\left( 11-1 \right)!} \right)\times \left( \dfrac{10!}{1!\left( 10-1 \right)!} \right) \\
 & =\left( \dfrac{11!}{10!} \right)\times \left( \dfrac{10!}{9!} \right) \\
\end{align}\]
As we know that$n!=1\times 2\times 3\times 4.........n$. So by using the same formula in above equation we will get;
\[\begin{align}
  & =\left( \dfrac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1} \right)\times \left( \dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1} \right) \\
 & =11\times 10 \\
 & =110 \\
\end{align}\]
So there will be 110 such combinations.
Total possible combination will be combinations we get from first case and combinations we got from second one i.e. $11+110=121$
Hence answer is 121 i.e. 121, three letter password combinations will be possible which will be symmetric.

Note: So moving ahead with the question, we had 11 symmetric elements. And we had to find out the three letter password that should also be symmetric. So we can directly say that there can be one type of arrangement in a password in which all three letters (which are symmetric) are the same. That can be AAA, HHH, III,.......... YYY. As these passwords will always be symmetric. So now there are 11 password combinations we got.
Other than this if we have two symmetric alphabets at the end and one asymmetric alphabet in the middle which is different to other two alphabets at the end, this combination will also be symmetric. For example; HAH, IAI, AHA, YWY,.......... AYA. So we can write it as a combination of, at the ends of 11 symmetric alphabet 1 alphabet and from the remaining 10 remaining symmetric alphabet we want any 1. So relating it with the concept of combination we can write it as$^{11}{{C}_{1}}{{\times }^{10}}{{C}_{1}}$. So we can write it as;
\[\begin{align}
  & =\left( \dfrac{11!}{1!\left( 11-1 \right)!} \right)\times \left( \dfrac{10!}{1!\left( 10-1 \right)!} \right) \\
 & =\left( \dfrac{11!}{10!} \right)\times \left( \dfrac{10!}{9!} \right) \\
\end{align}\]
As we know that$n!=1\times 2\times 3\times 4.........n$. So by using the same formula in above equation we will get;
\[\begin{align}
  & =\left( \dfrac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1} \right)\times \left( \dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1} \right) \\
 & =11\times 10 \\
 & =110 \\
\end{align}\]
So there will be 110 such combinations.
Total possible combination will be combinations we get from first case and combinations we got from second one i.e. $11+110=121$
Hence the answer is 121 i.e. 121, three letter password combinations will be possible which will be symmetric.