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
\(1-\frac{15}{24}=\frac{9}{24}=\frac{3}{8}\)

Here is my solution...

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P(no object is in the right place) = (# outcomes where no object is in right place)/(TOTAL # number of outcomes)

ALWAYS begin with the denominator.
To understand WHY we should start with the denominator, see our free video - http://www.gmatprepnow.com/module/gmat- ... /video/742

TOTAL # number of outcomes
We can arrange the 4 objects in 4! ways (=24 ways)

# outcomes where no object is in right place
Since, we know there are only 24 possible outcomes in TOTAL, it might not be a big deal just LIST the outcomes where no object is in right place.
We'll list the objects (1, 2, 3, and 4) in an order so that no number is in the correct place.

To begin, we'll place a 2 in the first position to get the following:
2134
2341
2413
There are 3 possible outcomes.

IF we place a 3 in the first position, we should also get 3 possible outcomes.
To confirm:
3142
3412
3421
Yep, 3 possible outcomes

IF we place a 4 in the first position, we should also get 4 possible outcomes.
To confirm:
4123
4312
4321
Yep, 3 possible outcomes

# of outcomes where no object is in right place = 3 + 3 + 3 = 9
---------------------------
So, P(no object is in the right place) = 9/24 = 3/8 = B
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I think you meant 4!=24 and not 16.

Good catch!
I have edited my response.

Cheers,
Brent
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Good catch!
I have edited my response.

Cheers,
Brent
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Hi Bunuel, I went through the first solution but I did not get this part, Please explain.

Thanks!!

Similar questions to practice:
https://gmatclub.com/forum/tanya-prepar ... 16819.html
http://gmatclub.com/forum/tanya-prepare ... 88626.html
https://gmatclub.com/forum/each-of-four ... 01553.html
http://gmatclub.com/forum/letter-arrang ... 84912.html
http://gmatclub.com/forum/micky-has-10- ... 13801.html
http://gmatclub.com/forum/5-letters-hav ... 89501.html
https://gmatclub.com/forum/5-letters-ha ... 14675.html

ALL POSSIBLE CASES: https://gmatclub.com/forum/letter-arran ... 84912.html
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This post is not for GMAT prep... It is for Maths lover http://oeis.org/wiki/Number_of_derangements
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Hi Brent

When you take the 1st number as 2. Is 2134 allowed? because in 2134, 3 falls in the 3rd place. Please help.

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