Aptitude Discussion

Q. |
Four different objects 1, 2, 3, 4 are distributed at random in four places marked 1, 2, 3, 4. What is the probability that none of the objects occupy the place corresponding to its number? |

✖ A. |
17/24 |

✔ B. |
3/8 |

✖ C. |
1/2 |

✖ D. |
5/8 |

**Solution:**

Option(**B**) is correct

First of all, if we IGNORE the condition about where the objects can be placed, we can arrange the 4 different objects in 4! ways (= 24 ways).

So, we now must determine HOW MANY of those 24 arrangements are such that no objects occupy the location corresponding to its number.

A quick way to do this is to LIST acceptable outcomes.

**IMPORTANT:** We'll list each arrangement so that the first number represents the object that goes to location #1, the second number represents the object that goes to location #2, and so on.

So, for example, 3421 represents object #3 in location #1, object #4 in location #2, object #2 in location #3, and object #1 in location #4.

Let's be systematic:

Arrangements where object #2 is in location #1

The possible arrangements where NO object is in the correct location are as follows:

**2**143, **2**341, **2**413

Total number of arrangements $= 3$

Arrangements where object #3 is in location #1

The possible arrangements where NO object is in the correct location are as follows:

**3**142, **3**412, **3**421

Total # of arrangements $= 3$

Arrangements where object #4 is in location #1

The possible arrangements where NO object is in the correct location are as follows:

**4**123, **4**312, **4**321

Total # of arrangements $= 3 $

Altogether, the number of arrangements where no object is in the correct location,

$= 3 + 3 + 3$

$= 9 $

So, $P(\text{no object in correct location}) = \dfrac{9}{24}$

$= \dfrac{3}{8}$

Thus, option (B) is the correct choice.

**Edit:** Based on **Vaibhav**'s comment, the solution has been updated.

**Edit 2:** For an alternate solution see **Kartik**'s comment

**Edit 3: Sheldon** has provided a way to reach the correct answer without even solving the question

**Edit 4: Suzen** has given an exhaustive solution.

**Edit 5:** For derangements formula, check **Himanshu Singh's** comment.

**Note:** For better understanding and different approaches do visit comment section.

**Gaurav Karnani**

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**Himanshu Singh**

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Please use this formula for derangements :

$!n = n!\left(1-\dfrac{1}{1!} + \dfrac{1}{2!} - \dfrac{1}{3!} + \dfrac{1}{4!} ... (-1)^n \times \dfrac{1}{n!} \right)$

**Kartik**

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Total number of ways of placing objects into places randomly,

$=4 \times 3 \times 2 \times 1=4!=24$

No. of ways of placing them such that only one of them gets correct place,

$\begin{align*}

= & 4 \text{ (ways of choosing the correctly} \\

& \text{ placed objects)} \times \\

& \times 2 \text{ (to place next object wrongly)} \\

& \times 1 \text{ (to place next object wrongly)} \\

& \times 1 \text{ (to place next object wrongly)}\\

= & \textbf{8}

\end{align*}$

Number of ways of placing exactly two objects correctly,

$\begin{align*}

= & ^4C_2 \text{ (to identify the two correctly placed} \\

& \text{ objects their arrangment are unique}\\

& \text{ for them to be correct)} \times \\

& 1 \text{ (to arrange next object wrongly)} \times \\

& 1 \text{ (to arrange next object wrongly)}\\

= & \textbf{6}

\end{align*}$

Now,

Number of ways to place exctly 3 objects correct = ways to place all objects correct $= \textbf{1}$

Therefore, number of ways to place all digits wrongly,

$=24-(8+6+1)$

$=24-15$

$=9$

Thus, Required probability,

$=\dfrac{9}{24}$

$=\dfrac{3}{8}$

**Raj**

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Since there's only one correct way of placing the objects as per the boxes- can't we work on the lines of using probability of an event happening = 1 - probability of that event not happening?

So therefore, can't the answer be 1 - (1/24) = 23/24?

That's a good idea, but it doesn't apply here.

If event A = NONE of the objects in the correct places, then the complement (event A NOT happening) will consist of any arrangement where some (perhaps all) of the objects are in their correct place(s).

**Vaibhav**

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but this is the case for 2 at first position. it can occupy position 3 and 4 as well . Similar case with other digits will be there.

You are right Vaibhav, updated the solution.

**Suzen**

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Here's the full list:

1234

1243

1423

1432

1324

1342

2134

2143 OK

2431

2413 OK

2341 OK

2314

3124

3142 OK (Close to 1000pi, but no correlation really, or is there? Perhaps the nth position is not n for all decimal places. OK, an interesting idea to explore)

3421 OK

3412 OK

3214

3241

4123 OK

4132

4321 OK

4312 OK

4213

4231

Clearly $P = \frac{9}{24} = \frac{3}{8}$

I know it's long winded, but I'm glad it matched!

**Tesla**

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Let a particular number (say) number 2 occupies position 1.

Then all possible arrangement are given as:

(2,1,3,4), (2,1,4,3), (2,3,4,1), (2,4,1,4), (2,4,1,3), (2,4,3,1).

Out of these six, three (2,1,3,4), (2,3,1,4), (2,4,3,1) are not acceptable because numbers 3 and 4 occupy the correct positions.

Required probability = 3/6 = 1/2

**Sheldon**

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I am not sure what difficulty level this would fall into.

If I see this question in the real exam, I would spend about a minute solving it.

After that, I would guess and move on. because it would be time consuming.

Looking at the question, it is more obvious that at-least 1 number would fall into its corresponding place.

So definitely, the probability that the number wouldn't fall into its numbered place should be less than half.

Looking at options:

A. 17/24

B. 3/8

C. 1/2

D. 5/8

Only option B is less than half. I would pick B and move on.

That's a great approach that allows you to minimize time spent on a question (that you feel is going nowhere) and maximize your guess.

Probability questions are perfect for this, because most people have a gut feeling about how likely something is. As you suggest, it does seem unlikely (probability less than 0.5) that every object would be out of place, so B is the perfect guess.

**Steve**

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I agree with sam and abhishek

**Abhishek**

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the correct answer will be $\dfrac{9}{24} =\dfrac{3}{8}$

out of

1234 1243 1324 1342 1423 1432

2341 2314 2143 2134 2413 2431

3124 3142 3412 3421 3214 3241

4123 4132 4213 4231 4312 4321

24 combinations

2341 2143 2413

3421 3412 3142

4123 4312 4321

are misplaced(dearranged)

So, the correct answer will be $\dfrac{9}{24}$ i.e. $\dfrac{3}{8}$. :)

I think Abhishek is correct...... answer should be 3/8

Total number of cases = 24

let us divide the case into categories where :

1) single number is in right position = 8 cases; two for each $1,2,3,4$

2) Two numbers are in right position = 6 cases $12,13,14,23,24,34$

3) No case for three digits as if three are in correct position fourth one will definitely be, so only one case of $1234$ .

total number of cases =15

Probability = $9/24$