The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:

Label the keys A, B, C, and D, such that key A fits the first lock, key B fits the second lock, and so on.  Each possible reassignment of the keys can then be seen as a rearrangement of the four letters.  For instance, the “word” BCAD would correspond to the reassigning key B to the first lock, key C to the second lock, key A to the third lock, and key D to the fourth lock.  In this particular case, only key D would fit its lock.

Thus, we should compute the number of anagrams of ABCD in which exactly two of the letters are in their original alphabetic positions.

There are at least two ways to compute this number:

1) Simply try listing the possibilities.  First place two letters in correct positions, then fill in the others.  The letters in their correct positions will be written in uppercase; letters out of position will be written in lowercase.

Correct letters     Anagram

A and B               ABdc
A and C               AdCb
A and D               AcbD
B and C               dBCa
B and D               cBaD
C and D               baCD

Notice that once you have chosen the two correct letters, the positions of the other two letters are fixed.  Thus, there are 6 possible rearrangements of the letters with exactly two in correct positions.

2) Apply combinatorics principles.  You need to choose the two correct letters out of four possible letters.  This means that the number of choices is the number of groups of 2 you can choose out of 4.  It doesn’t matter in what order you pick those two correct letters.  Notice that in the table above, we wrote “A and B” as one possibility—there are not TWO possibilities, “A and B” and “B and A.”  Thus, we have a situation in which order does not matter as we pick the two correct letters, so we write an anagram grid:

Alternatively, we can recognize this computation as “4 choose 2,” a combination.  Regardless, we write (4!)/(2!x2!) = 6 possibilities.

Whichever way we calculate the 6 possibilities, we can finish the problem in a straightforward manner at this point.  Since there are 4! = 24 total possible rearrangements of the 4 letters, the probability that exactly two keys fit their locks is 6/24 = 1/4.

The correct answer is C.

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