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

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.

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