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13 Oct 2015, 00:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
47% (02:32) correct
53%
(02:22)
wrong
based on 180
sessions
History
Date
Time
Result
Not Attempted Yet
Four different objects 1,2,3,4 are distributed at random in four places marked 1,2,3,4. What is probability that none of the objects occupy the place corresponding to it's number?
A) 17/24
B) 3/8
C) 1/2
D) 5/8
E) 1/3
www.GMATinsight.com
A) 17/24
B) 3/8
C) 1/2
D) 5/8
E) 1/3
www.GMATinsight.com
Updated on: 14 Oct 2015, 06:30
Originally posted by Marchewski on 13 Oct 2015, 17:17.
Last edited by Marchewski on 14 Oct 2015, 06:30, edited 1 time in total.
Last edited by Marchewski on 14 Oct 2015, 06:30, edited 1 time in total.
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Let me give it a try.
There are three different cases:
(1) Three correspond
(2) Two correspond
(3) One corresponds
(1) There is only \(1\) way for three to match. If 1,2,3 match, then 4 must be in its position too. Thus, all of them will be in their corresponding position.
(2) There are \(2C4=6\) ways for two to match. I thought about it this way: For each number pair, there is exactly one way for the other two numbers not to match their position. For example, if 1 and 2 match their position, then 3 and 4 must be in wrong positions. I can choose 6 different pairs out of 4 numbers.
(3) There are \(4 * 2 = 8\) ways for only one number to match its position. For example, there are only two ways for number 1 to be on its position while the other three are not: 1,3,4,2 and 1,4,2,3.
Hence, there are overall \(1 + 6 + 8 = 15\) different ways that at least one object corresponds to its number and \(4!=24\) ways to organize four objects.
The probability that no number occupies the place corresponding to it's number is therefore
\(1-\frac{15}{24}=\frac{9}{24}=\frac{3}{8}\)
There are three different cases:
(1) Three correspond
(2) Two correspond
(3) One corresponds
(1) There is only \(1\) way for three to match. If 1,2,3 match, then 4 must be in its position too. Thus, all of them will be in their corresponding position.
(2) There are \(2C4=6\) ways for two to match. I thought about it this way: For each number pair, there is exactly one way for the other two numbers not to match their position. For example, if 1 and 2 match their position, then 3 and 4 must be in wrong positions. I can choose 6 different pairs out of 4 numbers.
(3) There are \(4 * 2 = 8\) ways for only one number to match its position. For example, there are only two ways for number 1 to be on its position while the other three are not: 1,3,4,2 and 1,4,2,3.
Hence, there are overall \(1 + 6 + 8 = 15\) different ways that at least one object corresponds to its number and \(4!=24\) ways to organize four objects.
The probability that no number occupies the place corresponding to it's number is therefore
\(1-\frac{15}{24}=\frac{9}{24}=\frac{3}{8}\)
13 Oct 2015, 23:27
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Here is my solution...
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