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16 Sep 2014, 01:53
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C
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Each of four different locks has a matching key. The keys are randomly reassigned to the locks. What is the probability that exactly two of the keys fit the locks to which they are reassigned?
A. \(\frac{1}{8}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{4}\)
D. \(\frac{3}{8}\)
E. \(\frac{1}{2}\)
A. \(\frac{1}{8}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{4}\)
D. \(\frac{3}{8}\)
E. \(\frac{1}{2}\)
16 Sep 2014, 01:53
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Official Solution:
Each of four different locks has a matching key. The keys are randomly reassigned to the locks. What is the probability that exactly two of the keys fit the locks to which they are reassigned?
A. \(\frac{1}{8}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{4}\)
D. \(\frac{3}{8}\)
E. \(\frac{1}{2}\)
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. This method is much harder conceptually, and involves a couple of steps.
First, you need to choose which of the two letters out of the four possible letters are going to be "properly placed." This means that you count 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:
ABCD
YYNN
Alternatively, we can recognize this computation as "4 choose 2," a combination. Regardless, we write \(\frac{4!}{2!*2!} = 6\) possible choices of two "properly placed" letters.
Now that we have assigned two letters to the "proper placement" subset and the two leftover letters to the "improper placement" subset, we must consider how many ways each of those subsets of letters can actually be placed. The two letters that are to be properly placed can only be placed one way, since each letter has just one correct position. What about the improperly placed subset? Imagine that those two letters are A and B. Well, then, they must be placed as BA. There is exactly one way to "derange" two letters (that is, to place them in such a way that neither is in its original position).
Long story short, once we have chosen which 2 of the 4 letters are to be properly placed and which 2 are to be improperly placed, there is only one way to place the 4 letters. For instance, if we say that you will place A & C properly and B & D improperly, then the only arrangement that works is ADCB (written AdCb earlier).
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 \(\frac{6}{24} = \frac{1}{4}\).
Answer: C
Each of four different locks has a matching key. The keys are randomly reassigned to the locks. What is the probability that exactly two of the keys fit the locks to which they are reassigned?
A. \(\frac{1}{8}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{4}\)
D. \(\frac{3}{8}\)
E. \(\frac{1}{2}\)
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. This method is much harder conceptually, and involves a couple of steps.
First, you need to choose which of the two letters out of the four possible letters are going to be "properly placed." This means that you count 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:
ABCD
YYNN
Alternatively, we can recognize this computation as "4 choose 2," a combination. Regardless, we write \(\frac{4!}{2!*2!} = 6\) possible choices of two "properly placed" letters.
Now that we have assigned two letters to the "proper placement" subset and the two leftover letters to the "improper placement" subset, we must consider how many ways each of those subsets of letters can actually be placed. The two letters that are to be properly placed can only be placed one way, since each letter has just one correct position. What about the improperly placed subset? Imagine that those two letters are A and B. Well, then, they must be placed as BA. There is exactly one way to "derange" two letters (that is, to place them in such a way that neither is in its original position).
Long story short, once we have chosen which 2 of the 4 letters are to be properly placed and which 2 are to be improperly placed, there is only one way to place the 4 letters. For instance, if we say that you will place A & C properly and B & D improperly, then the only arrangement that works is ADCB (written AdCb earlier).
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 \(\frac{6}{24} = \frac{1}{4}\).
Answer: C
17 Feb 2017, 07:28
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Points to learn:
Dearrangement principle infers that
(i) 2 items may be dearranged in only 1 way
(i) 3 items may be dearranged in only 2 ways
Dearrangement principle infers that
(i) 2 items may be dearranged in only 1 way
(i) 3 items may be dearranged in only 2 ways
