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Problem - P & C

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Problem - P & C

by 6983manish » Wed Mar 30, 2011 7:47 pm
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. 11768
B. 12870
C. 13880
D. 8800
E. 14870
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by Anurag@Gurome » Wed Mar 30, 2011 8:20 pm
6983manish wrote: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. 11768
B. 12870
C. 13880
D. 8800
E. 14870
Possible 3 letter words = 26P3 = 26 * 25 * 24 = 15600
Since 11 are symmetric letters, so remaining 15 are asymmetric letters.
Possible 3 letter words with asymmetric letters = 15P3 = 2730
Possible words with at least one symmetric letter = 15600 - 2730 = 12870

The correct answer is B.
Last edited by Anurag@Gurome on Wed Mar 30, 2011 8:35 pm, edited 1 time in total.
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by manpsingh87 » Wed Mar 30, 2011 8:28 pm
6983manish wrote: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. 11768
B. 12870
C. 13880
D. 8800
E. 14870
here 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!;

similarly for case 2 we have 11C2*15C1*3!
and for case 3 11C3*3!;

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

hence B
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by force5 » Wed Mar 30, 2011 11:59 pm
yes perfect anurag... B it is.
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