# Difficult Permutation-Combination Solved QuestionAptitude Discussion

 Q. Each of the $11$ letters $A, H, I, M, O, T, U, V, W, X$ and $Z$ 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?
 ✖ A. $12,000$ ✔ B. $12,870$ ✖ C. $13,000$ ✖ D. $\text{None of these}$

Solution:
Option(B) is correct

As there are $26$ alphabets, we can choose a $3$ letter password from $11$ symmetric letters and $26-11=15$  from asymmetric letters.

There are three possible cases that will satisfy the condition of forming three letter passwords with at least $1$ symmetric letter.

Case 1:

1 symmetric and 2 asymmetric

$=({^{11}C_1} × {^{15}C_2})$

$=11 × \left(\dfrac{15 × 4}{1 × 2}\right)$

$=1155$

Case 2:

2 symmetric and 1 asymmetric

$=({^{11}C_2} × {^{15}C_1})$

$=\left(\dfrac{11 × 10}{1 × 2}\right) × 15$

$=825$

Case 3:

all 3 symmetric

$= {^{11}C_3}$

$=\left(\dfrac{11 × 10 × 9}{1 × 2 × 3}\right)$

$=165$

Hence, total possible cases,

$= \{\text{case (1)}+\text{case (2)}+\text{case (3)}\}× 3!$

$= [1155 + 825 + 165] × 6$

$= 2145 × 6$

$=\textbf{12,870}$

Edit: For an alternative method (using negation), check out comment from Bob.

## (9) Comment(s)

Nabin Dhakal
()

it's a computer password guys. Shouldn't we use permutations rather than combinations?

Mst Atul
()

searching comment for the same!!!

Shubham
()

It should be "Y" not "Z"..

Charu
()

what a mad rush to solve the question .. did anyone even realized that z is neither vertically symmetric (like T, W or M) nor horizontally symmetric (B, D, E) nor both vertically + horizontally symmetric like O.

Everyone just rushed to solve an inadequate question !! ha ha ha

Kudos, Mathematicians. You just proved why Physicists can be mathematicians but why Mathematician can't be anything except mathematician.

Charu
()

what a mad rush to solve the question .. did anyone even realized that z is neither vertically symmetric (like T, W or M) nor horizontally symmetric (B, D, E) nor both vertically + horizontally symmetric like O.

Everyone just rushed to solve an inadequate question !! ha ha ha

Kudos, Mathematicians. You just proved why Physician can be mathematicians but why Mathematician can't be anything except mathematician.

Alex
()

12870 it is

=>Total no. of possible 3 letter words - no. of possible words using only asymmetric words

=> 26 * 25 * 24 - 15 * 14 * 13

= 12870

Bob
()

There are a total of 11 symmetric letters, and therefore, 15 asymmetric letters.

Total number of words possible (no repetition):

$26*25*24 = 650*24 = 15600$

Total number of words possible with only asymmetric letters:

$15*14*13 = 210*13 = 2730$

Total number of words with at least one symmetric letter:

$15600 - 2730 = 12870$

Bisen
()

Kudos Bob, elegant and simple! Permutations and combinations are my weaker part and you do help me here a lot.

I approached from the usual side:

1) Combinations of types of letters:

AAA - 1

AAS - 3 (3C2 = 3)

SSA - 3 (3C2 = 3)

2) Permutations within each combination

AAA = 11 x 10 x 9 = 990

AAS = 15 x 14 x 11 = 2310

SSA = 11 x 10 X 15 = 1650

3) Multiply by number of combinations and add those

990 + 2310 x 3 + 1650 x 3 = 12,870

Juhi
()

Yes finding total possible way and removing the exception is the easy way to approach this kind of question.

good question.