Question

Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears the same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?

Answer 1

We can make the 3 letter password with 1 symmetric and 2 asymmetric letters(${{N}_{1}}$ ways) or 2 symmetric and 1 asymmetric letters (${{N}_{2}}$ ways) or all 3 symmetric letters (${{N}_{3}}$ ways). We can also arrange the letters in $3!$ ways. We find the required result as $\left( {{N}_{1}}+{{N}_{2}}+{{N}_{3}} \right)3!$ ways.

Complete step by step answer:
We know the rule of product from the fundamental of counting that if there are $m$ ways to do something and $n$ ways to other things then there are $m\times n$ ways to do both things. We also know that the selection of $r$ entities from $n$ unique entities is given by $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

We are given that the 11 letters A, H, I, M, O, T, U, V, W, X and Z are called symmetric letters because they appear the same when looked at in a mirror. Other letters in the alphabet are asymmetric letters. The total number of letters in English alphabet is 26. So the number of asymmetric letters are $26-11=15$.
We are asked to find the number of ways a three letter password can be made with at least one symmetric letter. We get three cases
Case-1: We select 1 symmetric letter out of 11 and the rest 2 asymmetric letters out 15. We can do it in say ${{N}_{1}}$ number of ways. So we have
${{N}_{1}}{{=}^{11}}{{C}_{1}}{{\times }^{15}}{{C}_{2}}=1155$
Case-1: We select 2 symmetric letters out of 11 and the rest 1 asymmetric letter1 out 15. We can do it in say ${{N}_{2}}$ number of ways. So we have
${{N}_{2}}{{=}^{11}}{{C}_{2}}{{\times }^{15}}{{C}_{1}}=825$
Case-1: We select all 3 symmetric letters for the password.. We can do it in say ${{N}_{3}}$ number of ways. So we have
${{N}_{1}}{{=}^{11}}{{C}_{3}}=165$
We can also arrange the three letters after selection in $3!=6$ number of ways. So we use of rule of product and find total number of ways $N$ is
$N=\left( {{N}_{1}}+{{N}_{2}}+{{N}_{3}} \right)\times 6=\left( 1155+825+165 \right)\times 6=2145\times 6=12870$

Note: We can also solve the problem using negation. We first the total number of ways we can select 3 letters out of 26 and arrange them that is $3!{{\times }^{26}}{{C}_{3}}$ ways. We then find a total number of ways we can only select asymmetric letters that is $3!{{\times }^{15}}{{C}_{3}}$ ways . The difference between them will give the required result.