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22 Aug 2016, 00:51
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D
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Difficulty:
75%
(hard)
Question Stats:
59% (02:41) correct
41%
(02:48)
wrong
based on 264
sessions
History
Date
Time
Result
Not Attempted Yet
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
A. 2,145
B. 6,435
C. 12,100
D. 12,870
E. 25,740
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
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
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
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.
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
