Latest Elitmus Aptitude Question SOLUTION: Q13.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 asymmetr

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Q13.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)12000
(B)12870
(C)13000
(D)None of these

Asked In Elitmus Mukesh Nayal (9 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

Mrudula Mrudu  (9 years ago)

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

Is this solution Helpfull? Yes (44) | No (3)  

saloni daga   (9 years ago)

three cases are possible:
1) 1 symmetric, 2 asymmetric
2) 2 symmetric, 1 asymmetric
3) 3 symmetric

1)from 11 symmetric letters 1 can be selected in 11C1 ways, and from the remaining 15 asymmetric letters two letters can be selected in 15C2 ways. now these 3 selected numbers can arrange themselves in 3! ways. so total no. of ways=11C1*15C2*3!;

case 2: 11C2*15C1*3!
case 3: 11C3*3!;

=> required no. of ways= 11C1*15C2*3!+11C2*15C1*3!+11C3*3!=12870;

Is this solution Helpfull? Yes (40) | No   

pankaj  (9 years ago)

11c1*15c2*3! + 15c1*11c2*3! +11c3*3! = 12870

Is this solution Helpfull? Yes (4) | No   

naga malleswara rao yaganti  (9 years ago)

total number of letters=26
total possible 3 letter passwords=26p3
total non symmetric letters=26-11=15
total passwords possible=15p3
three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter=(26p3-15p3)
correct me if i am wrong

Is this solution Helpfull? Yes (3) | No (2)  

sonu sahoo  (9 years ago)

ans is none of these , three cases are there,when only one symmetric letter, when two and in last case all three are symmetric letter,
11*15*14+11*10*15+11*10*9=4950

Is this solution Helpfull? Yes (2) | No (7)  

Mukesh Nayal  (9 years ago)

@ NAGA is correct and initially i got the same answer as @Sonu but it is wrong (still don't know how)

Is this solution Helpfull? Yes (1) | No (3)  

Gauraw Rawat  (8 years ago)

(26*25*24)-(15*14*13)=12870

Is this solution Helpfull? Yes (1) | No   

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