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

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Al, Bob, Cal and Don each own 1 hat. If the 4 hats are  [#permalink]

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New post 06 Nov 2018, 08:35
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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  [#permalink]

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New post 08 Nov 2018, 06:59
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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

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  [#permalink]

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New post 18 Nov 2018, 01:03
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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

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Al, Bob, Cal and Don each own 1 hat. If the 4 hats are  [#permalink]

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New post Updated on: 26 Nov 2018, 11:02
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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
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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.
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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are  [#permalink]

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New post 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  [#permalink]

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New post 07 Nov 2018, 08:14
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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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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.
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
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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are  [#permalink]

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New post 18 Nov 2018, 17:36
Shobhit7 wrote:
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


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.
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Re: Al, Bob, Cal and Don each own 1 hat. If the 4 hats are   [#permalink] 18 Nov 2018, 17:36
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