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Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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06 Nov 2018, 08:35
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34% (02:15) correct 66% (02:05) wrong based on 110 sessions
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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
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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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08 Nov 2018, 06:59
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 hatLet 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, Db, 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 outcomesWe 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
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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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06 Nov 2018, 12:23
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



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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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07 Nov 2018, 08:14
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
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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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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) Can you offer an alternative?



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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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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.
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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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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.



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Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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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)
But looking for better approach



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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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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!.
Probability: 9/24 = 3/8 Ans D



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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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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 Can you pls clarify



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Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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17 Nov 2018, 21:20
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. Ab, Bc, ca Ac, Ba, Cb # 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



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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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18 Nov 2018, 01:03
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



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Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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Updated on: 26 Nov 2018, 11:02
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! ) = 124+1 = 9. Total possible arrangements = 4! = 24. P(no hat is in the correct position) = 9/24= 3/8. IMO D
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Last edited by warrior1991 on 26 Nov 2018, 11:02, edited 1 time in total.



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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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18 Nov 2018, 17:36
Shobhit7 wrote: Hi AKY13Case 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 deviceSeems 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.



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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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16 Dec 2019, 12:40
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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are
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