It took me a while to figure this out; first thinking it could be done with permutations. Alas, I wasn't successful in that route.
I must admit, it helped to know what the ans was in advance (indicated by square box around D).
Here we go...
I drew a simple drawing, with letters in top row, matching addresses on bottom row:
ABCD
1234
Instance 1: Assume letter A is paired with envelope 1 and the rest are mismatched.
(1/4) - First, out of 4 letters, 1 is correct
(2/3) - Second, out of remaining 3 letters, 2 letters (only 2 and not all 3!) will satisfy the condition that second envelope is mismatched.
(1/2) - Third, out of remaining 2 letters, 1 letter is mismatch-able
(1/1) - Finally, only one letter and one envelope remains.
Above only accounts for one instance and does not account for letter B correctly paired with envelope 2, and so forth.. There are FOUR instances altogether.
Thus, 4 * (1/4) * (2/3) * (1/2) * (1/1) = 1/3 ... ans D
Gmat Prep Question (Prob) Another one..
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