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.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Does GMAT RC seem like an uphill battle? e-GMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days.
Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to pre-think assumptions and solve the most challenging questions in less than 2 minutes.
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
Show Tags
06 Nov 2018, 08:35
1
Top Contributor
17
00:00
A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
34% (02:05) correct 66% (01:58) wrong based on 130 sessions
HideShow timer Statistics
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?
Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
Show Tags
08 Nov 2018, 06:59
Top Contributor
1
GMATPrepNow wrote:
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
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
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!
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
Show Tags
Updated on: 26 Nov 2018, 11:02
1
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
_________________
-------------------------------------------------------------------------------------------------------------------------- All the Gods, All the Heavens, and All the Hells lie within you.
Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
Show Tags
07 Nov 2018, 08:14
Top Contributor
1
jsistare wrote:
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?
Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
Show Tags
07 Nov 2018, 09:01
GMATPrepNow wrote:
jsistare wrote:
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)
Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
Show Tags
07 Nov 2018, 09:22
jsistare wrote:
GMATPrepNow wrote:
jsistare wrote:
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?
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.
_________________
The importance of mindset on the GMAT - 640 to 690 to 740 (Q49 V42) : https://bit.ly/2R1WaK5
Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
Show Tags
07 Nov 2018, 09:46
GMATPrepNow wrote:
jsistare wrote:
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
Could you please give the solution, I'm stuck on it I always get 3/4.
Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
Show Tags
07 Nov 2018, 23:58
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)
Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
Show Tags
08 Nov 2018, 12:29
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!.
Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
Show Tags
17 Nov 2018, 20:49
Shobhit7 wrote:
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
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
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
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
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.
gmatclubot
Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
[#permalink]
18 Nov 2018, 17:36
close
34% (02:05) correct 
