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Bunuel
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Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same 22 Aug 2016, 00:51
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60% (02:40) correct 40% (02:48) wrong based on 291 sessions

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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
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D
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udyans123
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Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same 22 Aug 2016, 08:09
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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
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GMAT Focus 1: 735 Q90 V89 DI81
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Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same 09 Nov 2020, 03:21
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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.
Show more


The main flaw is you have taken an approach that is lengthy, calculations intensive and error prone.

Now the flaw in the calculations.
If you calculate 15*14, it already includes the arrangement so you cannot multiply by 6.
1) 1 S and 2 AS = 11C1*15C2*3!=\(11*\frac{15*14}{2}*6\)
2) 1AS and 2S = 11C2*15*6=\(\frac{11*10}{2}\)*15*6
3) 3S=11*10*9
Add and you will get the answer. => 6930 + 4950 + 990 = 12,870

The better way would be
All ways-(ways none of S is there)
All ways =26*25*24
Ways where none is symmetric =15*14*13
Our answer, that is at least one is symmetric = 26*25*24-15*14*13=15600-2730=12870
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Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same 22 Aug 2016, 06:34
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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
Show more


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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Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same 23 Apr 2017, 15:09
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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
Show more

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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Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same 07 May 2017, 22:58
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A quick observation about the question.
Z is not a symmetric letter by this definition.
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Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same 27 Aug 2017, 20:58
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this tests the ability to calculate fast and accurate
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Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same 08 Nov 2020, 08:24
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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
Show more

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 :)

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Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same 04 Dec 2020, 03:07
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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
Show more

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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Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same 22 Feb 2025, 04:06
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