Problem - P & C - The Beat The GMAT Forum - Expert GMAT Help & MBA Admissions Advice

Problem - P & C

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6983manish: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Mon Mar 07, 2011 12:27 am; Thanked: 6 times

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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Source: Beat The GMAT — Problem Solving | Wed Mar 30, 2011 7:47 pm

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

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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force5: Legendary Member; Posts: 582; Joined: Tue Mar 08, 2011 12:48 amFollowed by: 6 members; Thanked: 61 times; GMAT Score:740

by force5 » Wed Mar 30, 2011 11:59 pm

yes perfect anurag... B it is.

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