Self reflected passwords

Probability Level 4

Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Y appears 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?

The answer is 12870.

3 solutions

Vincent Miller Moral
Jul 22, 2015

Case 1: 1 symmetric letter; 2 asymmetric
(11)(15)(14) = 2310
11 ~ no. of ways to choose a symmetric
15 and 14 ~ no. of ways to choose 2 asymmetric

2310 --> 2310*3 = 6930

Case 2: 2 symmetric; 1 asymmetric
(11)(10)(15) = 1650
1650 --> 1650*3 = 4950

Case 3: All symmetric
(11)(10)(9) = 990

Answer = 6930 + 4950 + 990 = 12870

Vincent Tandya
Jul 22, 2015

One might attempt to solve this problem by counting how many possible passwords contain exactly one, two, and three symmetric letters, and then sum it. However, complementary counting reduces the amount of counting needed. In this problem, we could just count how many passwords have no more than one symmetric letter (which means all the three letters are asymmetric), and how many passwords are able to be arranged, neglecting any restrictions.

Note that there are $\frac{15!}{(15-3)!}$ ways to arrange three distinct asymmetric letters out of $16$ asymmetric letters available, while there are $\frac{26!}{(26-3)!}$ ways to arrange three distinct letters from $26$ letters in the alphabet.

So, there are $\frac{26!}{23!} - \frac{15!}{12!} = 26 \times 25 \times 24 - 15 \times 14 \times 13 = \fbox{12870}$ possible passwords consisting theee letters, including at least one symmetric letter.

Saya Suka
May 7, 2021

n(at least one symmetric letter in 3-letter codes)
= n(all possible 3-letter codes) – n(all asymmetric letters in 3-letter codes)
= ( 26 × 25 × 24 ) – ( 15 × 14 × 13 )
= 12870

Self reflected passwords

The answer is 12870.

3 solutions

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