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Bunuel
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. 2,145
B. 6,435
C. 12,100
D. 12,870
E. 25,740


Since we are given atleast one symmetric letter in the three letter word we can take the following cases

1. All three
2. One symmetry and other two non
3. Two symmetry and one non
4. All the three letters can be arranged in 6 ways

( 11c3 + 11c1 * 15c2 + 11c2 * 15c1 ) * 6

( 11*10*9/ 6 + 11 * 15 * 14 / 2 + 11*10 / 2 * 15 ) * 6

12870

IMO option D is the correct answer..

OA please...will correct if I missed anything..
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udyans123
Total 3 letter passwords = 26 * 25 * 24 = 15600

Total letters = 26 = symmetrical letters (11) + asymmetrical letters (x) => x = 15

Total no. of 3 letters passwords using asymmetrical letters = 15 * 14 * 13 = 2730 (no repetition allowed)

Hence, no. of 3 letter passwords using atleast one symmetrical letter = 15600 - 2730 = 12870 ---> Option D

I am still confusing between kCn and kPn, I thought kPn should be applied to this question, can you explain the concepts for me?
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A quick observation about the question.
Z is not a symmetric letter by this definition.
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this tests the ability to calculate fast and accurate
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Bunuel
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. 2,145
B. 6,435
C. 12,100
D. 12,870
E. 25,740

Hi Bunuel ,

What is wrong in my approach?

Symmetric letters (S)=11
Asymmetric letters (A)=15

3 case:

1 S and 2 AS = (11×15×14)×6 [6 is the number of ways] = 13860
OR
2 S and 1 AS = (11×10×15)×6 = 9900
OR
3 S = (11×10×9)×6 = 5940

Therefore, total ways = 29700

It would be great if you could help me out with the correct solution.

Tagging others just in case
chetan2u GMATinsight IanStewart ScottTargetTestPrep yashikaaggarwal VeritasKarishma


Thank you :)

Posted from my mobile device
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Nups1324
Bunuel
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. 2,145
B. 6,435
C. 12,100
D. 12,870
E. 25,740


What is wrong in my approach?

Symmetric letters (S)=11
Asymmetric letters (A)=15

3 case:

1 S and 2 AS = (11×15×14)×6 [6 is the number of ways] = 13860
OR
2 S and 1 AS = (11×10×15)×6 = 9900
OR
3 S = (11×10×9)×6 = 5940

Therefore, total ways = 29700

It would be great if you could help me out with the correct solution.



Thank you :)

Posted from my mobile device

Solution:

The mistake in your calculation is multiplying all three cases by 6. Notice that one S and two A can be arranged in 3!/2! = 3 ways, not 6. In fact, we can list all the possibilities of having one S and two A in a password: SAA, ASA and AAS. The same goes for two S and one A, it can also be arranged in three ways. For the case of three S, there is only one way to arrange SSS. Thus, you should multiply 11 * 15 * 14 and 11 * 10 * 15 by 3 and add 11 * 10 * 9 to the sum. You will get 12,870.
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