Last visit was: 18 Nov 2025, 22:35 It is currently 18 Nov 2025, 22:35
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
805+ Level|   Probability|      
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,739
Own Kudos:
35,331
 [46]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,739
Kudos: 35,331
 [46]
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 06 Nov 2018, 08:35
1
Kudos
Add Kudos
45
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
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:

37% (02:09) correct 63% (02:01) wrong based on 244 sessions

History

Date
Time
Result
Not Attempted Yet
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are randomly distributed so that each man receives exactly 1 hat, what is the probability that no one receives his own hat?

A) 1/8
B) 1/4
C) 1/3
D) 3/8
E) 1/2
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,739
Own Kudos:
35,331
 [4]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,739
Kudos: 35,331
 [4]
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 08 Nov 2018, 06:59
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
GMATPrepNow
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are randomly distributed so that each man receives exactly 1 hat, what is the probability that no one receives his own hat?

A) 1/8
B) 1/4
C) 1/3
D) 3/8
E) 1/2
Show more

I created this question to highlight many students' tendency to avoid listing and counting as a possible approach.
As you'll see, the approach is probably the fastest approach.

P(no one receives his own hat) = (# of outcomes in which no one receives his own hat)/(TOTAL number of outcomes)

# of outcomes in which no one receives his own hat
Let a, b, c and d represent the hats owned by Al (A), Bob (B), Cal (C) and Don (D)
Let's systematically list the HATS to be paired up with A, B, C, and D
A, B, C, D
b, a, d, c
b, c, d, a
b, d, a, c

c, a, d, b
c, d, a, b
c, d, b, a

d, a, b, c
d, c, a, b
d, c, b, a

So, there are 9 outcomes in which one receives his own hat


TOTAL number of outcomes
We can arrange n unique objects in n! ways
So, we can arrange the 4 hats in 4! ways (= 24 ways)
So, there are 24 possible outcomes

P(no one receives his own hat) = 9/24 = 3/8

Answer: D

Cheers,
Brent
General Discussion
avatar
jsistare
avatar
Joined: 15 Jan 2018
Last visit: 17 Dec 2019
Posts: 3
Own Kudos:
2
 [2]
Posts: 3
Kudos: 2
 [2]
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 06 Nov 2018, 12:23
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
4 hats can be distributed to 4 recipients in 4! ways --> 24 combinations

# of ways to assign hats so that nobody receives their own hat:
slot method: 3 x 2 x 1 x 1 = 3! ways --> 6 favorable combinations

(6 favorable) / (24 total) = 1/4
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,739
Own Kudos:
35,331
 [1]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,739
Kudos: 35,331
 [1]
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 07 Nov 2018, 08:14
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
jsistare
4 hats can be distributed to 4 recipients in 4! ways --> 24 combinations

# of ways to assign hats so that nobody receives their own hat:
slot method: 3 x 2 x 1 x 1 = 3! ways --> 6 favorable combinations

(6 favorable) / (24 total) = 1/4
Show more

Be careful; the correct answer is not 1/4.
When you use the slot method (aka the Fundamental Counting Principle), you must clearly define what each slot/stage represents.
For example, you have a 3 in the first slot, what does that represent?

Cheers,
Brent
avatar
jsistare
avatar
Joined: 15 Jan 2018
Last visit: 17 Dec 2019
Posts: 3
Own Kudos:
2
Posts: 3
Kudos: 2
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 07 Nov 2018, 09:01
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATPrepNow
jsistare
4 hats can be distributed to 4 recipients in 4! ways --> 24 combinations

# of ways to assign hats so that nobody receives their own hat:
slot method: 3 x 2 x 1 x 1 = 3! ways --> 6 favorable combinations

(6 favorable) / (24 total) = 1/4

Be careful; the correct answer is not 1/4.
When you use the slot method (aka the Fundamental Counting Principle), you must clearly define what each slot/stage represents.
For example, you have a 3 in the first slot, what does that represent?

Cheers,
Brent
Show more

The intent of the slot method was that each slot represents a hat, and the numbers expressing how many people they can be assigned to without anybody getting their own hat.

3 (first hat can go to 3 people) x 2 (second hat can only go to 2 people) x 1 (third hat can only go to one person) x 1 (only person left)

Can you offer an alternative?
User avatar
shaarang
User avatar
Joined: 06 Sep 2018
Last visit: 06 Apr 2021
Posts: 72
Own Kudos:
386
 [1]
Given Kudos: 51
Location: India
Concentration: Finance, Entrepreneurship
Schools: Kellogg '22 (D) Tepper '22 (WA) Duke '22 (WA) Kenan-Flagler '22 (M$) Kelley '22 (A$) McDonough '22 (A$) Anderson '22 (D)
GMAT 1: 740 Q49 V42
GPA: 4
WE:Analyst (Finance: Investment Banking)
Products:
My Reviews
Schools: Kellogg '22 (D) Tepper '22 (WA) Duke '22 (WA) Kenan-Flagler '22 (M$) Kelley '22 (A$) McDonough '22 (A$) Anderson '22 (D)
GMAT 1: 740 Q49 V42
Posts: 72
Kudos: 386
 [1]
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 07 Nov 2018, 09:22
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jsistare
GMATPrepNow
jsistare
4 hats can be distributed to 4 recipients in 4! ways --> 24 combinations

# of ways to assign hats so that nobody receives their own hat:
slot method: 3 x 2 x 1 x 1 = 3! ways --> 6 favorable combinations

(6 favorable) / (24 total) = 1/4

Be careful; the correct answer is not 1/4.
When you use the slot method (aka the Fundamental Counting Principle), you must clearly define what each slot/stage represents.
For example, you have a 3 in the first slot, what does that represent?

Cheers,
Brent

The intent of the slot method was that each slot represents a hat, and the numbers expressing how many people they can be assigned to without anybody getting their own hat.

3 (first hat can go to 3 people) x 2 (second hat can only go to 2 people) x 1 (third hat can only go to one person) x 1 (only person left)

Can you offer an alternative?
Show more

This method is wrong because:

A can choose his hat 3 ways

BUT, if A chooses B's hat,

B can now choose his 3 ways too (instead of 2).

The logic repeats for C and D.
avatar
jorgetomas9
avatar
Joined: 19 Aug 2018
Last visit: 01 Dec 2024
Posts: 29
Own Kudos:
7
Given Kudos: 35
Posts: 29
Kudos: 7
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 07 Nov 2018, 09:46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATPrepNow
jsistare
4 hats can be distributed to 4 recipients in 4! ways --> 24 combinations

# of ways to assign hats so that nobody receives their own hat:
slot method: 3 x 2 x 1 x 1 = 3! ways --> 6 favorable combinations

(6 favorable) / (24 total) = 1/4

Be careful; the correct answer is not 1/4.
When you use the slot method (aka the Fundamental Counting Principle), you must clearly define what each slot/stage represents.
For example, you have a 3 in the first slot, what does that represent?

Cheers,
Brent
Show more
Could you please give the solution, I'm stuck on it I always get 3/4.
User avatar
hellosanthosh2k2
User avatar
Joined: 02 Apr 2014
Last visit: 07 Dec 2020
Posts: 361
Own Kudos:
597
Given Kudos: 1,227
Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34
GPA: 3.5
Schools: XLRI"20
GMAT 1: 700 Q50 V34
Posts: 361
Kudos: 597
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 07 Nov 2018, 23:58
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I used this approach:
not a better approach, especially, if there were more than 4 persons,
nevertheless, this is how i solved it:

P(no person receives own hat) = 1 - P(one or more person received own hat)

Case 1: One person received own hat,
Let A be that person, so among, B,C,D, the possible combination of none receiving own hat is : D B C, C D B = 2 combinations
similarly we can extend that single person receiving own hat to B, C, D, so total combo = 4 * 2 = 8

Case 2: two persons receive own hat
Let A,B receive own hat, then among C,D, only combo of person not receiving own hat is 1
Similiary we could hav chosen (AC, AD,.... = 4C2 combinations of right persons), so total combinations = 4C2 * 1 = 6

Case 3: three persons receive own hat, well this also means that 4th person also receive own hat, only one combo possible

So total combinations = 8 + 6 + 1 = 15

Total ways of distributing hats = 4! = 24

P(no person receives own hat) = 1 - P(one or more person received own hat)
= 1 - (15/24) = 1 - (5/8) = 3/8 = (D)

But looking for better approach
avatar
Shobhit7
avatar
Joined: 01 Feb 2017
Last visit: 29 Apr 2021
Posts: 240
Own Kudos:
426
Given Kudos: 148
Posts: 240
Kudos: 426
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 08 Nov 2018, 12:29
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Four Persons: A, B, C, D
Their four hats respectively: a, b, c, d

For A: Three possibilities with hats: b or c or d.
If A gets hat b, then B has 3 options to choose from a,c or d. Hence, C and D will be left with 1 option each i.e non c hat for C and non d hat for D. So, total options: 1*3*1*1= 3

Similarly, there are 3 options each while choosing hat c and hat d for Person A.
Therefore, there are 9 favorable combinations and total possible combinations are 4!.

Probability: 9/24 = 3/8
Ans D
User avatar
AKY13
User avatar
Joined: 29 Sep 2016
Last visit: 01 Nov 2019
Posts: 84
Own Kudos:
24
Given Kudos: 40
Posts: 84
Kudos: 24
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 17 Nov 2018, 20:49
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Shobhit7
Four Persons: A, B, C, D
Their four hats respectively: a, b, c, d

For A: Three possibilities with hats: b or c or d.
If A gets hat b, then B has 3 options to choose from a,c or d. Hence, C and D will be left with 1 option each i.e non c hat for C and non d hat for D. So, total options: 1*3*1*1= 3

Similarly, there are 3 options each while choosing hat c and hat d for Person A.
Therefore, there are 9 favorable combinations and total possible combinations are 4!.

Probability: 9/24 = 3/8
Ans D
Show more

Hi Shobhit7,
I differ here. I also wanted to do this way but this method rules out possibility of other combinations. Why didn't you consider them.
e.g. If A selects c or d - it will have 2 favorable options & then B will have only 2 options (a or c/d)
Similarly if A selects b, B will have 3 options from a, c & d

Can you pls clarify
User avatar
AKY13
User avatar
Joined: 29 Sep 2016
Last visit: 01 Nov 2019
Posts: 84
Own Kudos:
24
Given Kudos: 40
Posts: 84
Kudos: 24
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 17 Nov 2018, 21:20
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I agree with Hellosanthosh2k2's approach

Mine is same; the way of expression is different as I found it convenient -

P(all exchange incorrectly) = 1 - P(3 persons interchanging incorrectly + 2 persons exchanging hats + no interchange/exchange)


3 persons interchanging hats
A'B'C'D
# of ways 3 dashes can be placed at 4 places = 4C3 = 4 ways
Within dashes say A'B'C', # of ways they can interchange incorrectly = 2
e.g.
A-b, B-c, c-a
A-c, B-a, C-b
# of ways = 4 x 2 = 8

2 persons exchanging hats
A'B'CD
# of ways 2 dashes(') can be placed at 4 places = 4C2 = 6 ways
A' B' C D / A B' C' D / A B C' D' / A' B C D' / A B' C D'/ A' B C' D


No Interchange/exchange
# of ways 4 persons wear correctly = 1

Submission gives 6+8+1 = 15

TOTAL # of ways 4! = 24

1 - (15/24) = 9/24 = 3/8
avatar
Shobhit7
avatar
Joined: 01 Feb 2017
Last visit: 29 Apr 2021
Posts: 240
Own Kudos:
426
 [4]
Given Kudos: 148
Posts: 240
Kudos: 426
 [4]
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 18 Nov 2018, 01:03
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi AKY13

"Detailed calculations"

Person A has 3 options: b,c,d.
Now, I have split and considered these options as 3 separate cases.
Reason: In slot method, we must first address the slot with maximum constraints. And, in each case, this slot is different. Here's how:

Case 1: Person A gets hat b.
In this case, next slot/person to consider should be the one whose hat is already taken i.e. Person B.
Person B has 3 options to choose from: a,c,d: 3C1
Now, irrespective of whatever Person B chooses from above three, Person C and D are each left with one option each.
So, total options in case 1: 1*3*1*1= 3

Case 2: Person A gets hat c.
In this case, next slot/person to consider should be the one whose hat is already taken i.e. Person C.
Person C has 3 options to choose from: a,b,d: 3C1
Now, irrespective of whatever Person C chooses from above three, Person B and D are each left with one option each.
So, total options in case 1: 1*3*1*1= 3

Case 3: Person A gets hat d.
In this case, next slot/person to consider should be the one whose hat is already taken i.e. Person D.
Person D has 3 options to choose from: a,b,c: 3C1
Now, irrespective of whatever Person D chooses from above three, Person B and C are each left with one option each.
So, total options in case 1: 1*3*1*1= 3

Sum of favorable outcomes from above three cases: 9
Total possible outcomes: 4!

Probability: 9/24= 3/8
Ans D

Posted from my mobile device
User avatar
warrior1991
User avatar
Joined: 03 Mar 2017
Last visit: 03 Feb 2022
Posts: 573
Own Kudos:
437
 [4]
Given Kudos: 596
Location: India
Concentration: Operations, Technology
Products:
My Reviews
Posts: 573
Kudos: 437
 [4]
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are Updated on: 26 Nov 2018, 11:02
Originally posted by warrior1991 on 18 Nov 2018, 02:40.
Last edited by warrior1991 on 26 Nov 2018, 11:02, edited 1 time in total.
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Number of derangement = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)

Given 4 hats A, B, C, and D
Total number of derangements = 4! (1/2! - 1/3! + 1/4! ) = 12-4+1 = 9.
Total possible arrangements = 4! = 24.
P(no hat is in the correct position) = 9/24= 3/8.

IMO D
User avatar
AKY13
User avatar
Joined: 29 Sep 2016
Last visit: 01 Nov 2019
Posts: 84
Own Kudos:
24
 [1]
Given Kudos: 40
Posts: 84
Kudos: 24
 [1]
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 18 Nov 2018, 17:36
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Shobhit7
Hi AKY13

Case 1: Person A gets hat b.
In this case, next slot/person to consider should be the one whose hat is already taken i.e. Person B.
Person B has 3 options to choose from: a,c,d: 3C1
Now, irrespective of whatever Person B chooses from above three, Person C and D are each left with one option each.
So, total options in case 1: 1*3*1*1= 3

Posted from my mobile device
Show more

Seems better but I still have one query.
In the above quoted part, you have mentioned
Now, irrespective of whatever Person B chooses from above three, Person C and D are each left with one option each.

If B chooses hat c, C will have 2 options (a&d) & then D will have one option.
User avatar
Regor60
User avatar
Joined: 21 Nov 2021
Last visit: 17 Nov 2025
Posts: 528
Own Kudos:
383
Given Kudos: 459
Posts: 528
Kudos: 383
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 13 Aug 2023, 13:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
4!=24 ways to distribute the hats

Answer will be:

[24-ways all get their hats-
ways 2 " - ways 1 "]/24

Ways all get their hats: 1

Ways 2 get their hats = 4!/2!2! = 6 times ? = even number

Ways 1 gets hat = 4 ways to select that person times ? = even number

So the total ways will be

24-1-2 even numbers = odd number

The only answer with an odd numerator when its denominator is normalized to 24 is 3/8, D.

Posted from my mobile device
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,584
Own Kudos:
1,079
Posts: 38,584
Kudos: 1,079
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are 15 Aug 2024, 07:36
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Bunuel
Math Expert
105356 posts
Krunaal
Tuck School Moderator
805 posts