Bunuel wrote:

A three-letter code is formed using the letters A-L, such that no letter is used more than once. What is the probability that the code will have a string of three consecutive letters (e.g. A-B-C, F-E-D)?

(A) 1/55

(B) 1/66

(C) 2/17

(D) 1/110

(E) 2/55

Kudos for correct solution.

Now let us arrange three letter code in the form (A-B-C) using the letters from A to L

i.e we have = ABC, BCD, CDE, DEF, EFG, FGH, GHI, HIJ, IJK, JKL

i.e we have 10 options.

Similarly we can write the three letter code in the form (F-E-D) i.e in opposite direction and we still have 10 options.

Therefore we have a total of 20 options, which is the favourable outcomes

Now Total outcomes = 12 * 11 *10 (Letters from A to L is 12 and here order matters, so we cannot use combination )

therefore

Probability =\frac{No. of favorable outcomes}{Total no. of outcomes}or

Probability = \frac{20}{(12*11*10}) = \frac{20}{1320} = \frac{1}{66}
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