GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Oct 2018, 10:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# S99-01

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50002

### Show Tags

16 Sep 2014, 01:53
1
4
00:00

Difficulty:

85% (hard)

Question Stats:

41% (01:53) correct 59% (01:37) wrong based on 87 sessions

### HideShow timer Statistics

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. $$\frac{1}{2}$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 50002

### Show Tags

16 Sep 2014, 01:53
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 &amp; C properly and B &amp; 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 $$\frac{6}{24} = \frac{1}{4}$$.

_________________
Manager
Joined: 31 Jan 2017
Posts: 58
Location: India
GMAT 1: 680 Q49 V34
GPA: 4
WE: Project Management (Energy and Utilities)

### Show Tags

17 Feb 2017, 07:28
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
_________________

__________________________________
Kindly press "+1 Kudos" if the post helped

Intern
Joined: 04 Aug 2018
Posts: 6

### Show Tags

30 Sep 2018, 09:08
simpler way : --
1) total number of ways 4 keys can be assigned randomly is 24 i.e 4!
2) exactly two key matches the locks :- selection of 2 out of four, i.e. any two key matches to assigned locks..4c2..i.e 6

Locks :- A B C D
keys :- e f g h

cases:- ef, eg, eh, fg, fh , gh.

req probability = 6/24 = 1/4
Re: S99-01 &nbs [#permalink] 30 Sep 2018, 09:08
Display posts from previous: Sort by

# S99-01

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.