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

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;
10 years agoHelpfull: Yes(40) No(0)
11c1*15c2*3! + 15c1*11c2*3! +11c3*3! = 12870
10 years agoHelpfull: Yes(4) No(0)
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
10 years agoHelpfull: Yes(3) No(2)
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
10 years agoHelpfull: Yes(2) No(7)
@ NAGA is correct and initially i got the same answer as @Sonu but it is wrong (still don't know how)
10 years agoHelpfull: Yes(1) No(3)
(26*25*24)-(15*14*13)=12870
9 years agoHelpfull: Yes(1) No(0)

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