Gmat Prep Question (Prob) Another one..

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dferm: Master | Next Rank: 500 Posts; Posts: 446; Joined: Thu Jul 26, 2007 1:07 pm; Thanked: 6 times

by dferm » Tue Jun 24, 2008 10:26 am

Can someone throw some light on this one?

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Source: Beat The GMAT — Problem Solving | Tue Jun 24, 2008 10:26 am

beeparoo: Master | Next Rank: 500 Posts; Posts: 200; Joined: Sun Jun 17, 2007 10:46 am; Location: Canada; Thanked: 9 times

by beeparoo » Tue Jun 24, 2008 4:28 pm

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

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