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There were 6 members in Mr. George's family: his wife, 2 daughters, on

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There were 6 members in Mr. George's family: his wife, 2 daughters, on  [#permalink]

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New post 05 Mar 2019, 02:37
6
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

48% (02:25) correct 52% (02:29) wrong based on 86 sessions

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There were 6 members in Mr. George's family: his wife, 2 daughters, one son and mother and himself. They had 6 cars out of which 2 were identical cars of brand A, one car of brand B, third was of brand C, one was of brand D& the sixth car was brand of E. If all family members want to use separate cars, in how many ways can they choose if Mr. George's son refuged to take the car of brand A?

A. 180
B. 200
C. 240
D. 360
E. 480
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Re: There were 6 members in Mr. George's family: his wife, 2 daughters, on  [#permalink]

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New post 05 Mar 2019, 03:49
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his son can select the remaining four cars in 4 ways
and out of the remaining five cars 2 are identical and it can be done in 5!/2! ways.
So total number of ways in which he can select is 4*5!/2! = 4*120/2 =4*60 = 240
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Re: There were 6 members in Mr. George's family: his wife, 2 daughters, on  [#permalink]

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New post 05 Mar 2019, 03:49
his son can select the remaining four cars in 4 ways
and out of the remaining five cars 2 are identical and it can be done in 5!/2! ways.
So total number of ways in which he can select is 4*5!/2! = 4*120/2 =4*60 = 240
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There were 6 members in Mr. George's family: his wife, 2 daughters, on  [#permalink]

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New post 26 Jun 2019, 03:17
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it can be done in a easy way by alloting the son 'A' car which he does not want.
total no. of cases = 6!/2! as two cars are identical (dividing it by 2!)
subtract the case when son has 'A' car = 5! ( allot A car to son and remaining 5 members 5! ways
= 6!/2! - 5! = 240 ways. ( no. of ways they can choose a car if son refused to have 'A' car.)
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Re: There were 6 members in Mr. George's family: his wife, 2 daughters, on  [#permalink]

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New post 27 Jun 2019, 10:19
Hello I know my logic is wrong but i would be helpful if anyone can please clear concept/question understanding:

Constraint 1 : No family member can have same CAR hence George gets CAR A
So remaining combination 5! for 5 members
Constraint 2: SON will not use car A
So let's give SON car A and see possible combinations 4!
Remaining cases= 5!-4!= Number of cases when George has CAR A and SON does not have car A= 96

I know my solution is wrong, please point out my flaw
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Re: There were 6 members in Mr. George's family: his wife, 2 daughters, on  [#permalink]

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New post 27 Jun 2019, 10:31
Hello Anurag,
1st find total no. Of cases - 6!/2! (2! For identical cars)
Now son don't want A car than allot 'A' to him - remaining car 5 and 5 people left - arrangements 5!
Cases when son will not have 'A' car = total - when he has A car
6!/2! - 5! = 240

Posted from my mobile device
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Re: There were 6 members in Mr. George's family: his wife, 2 daughters, on  [#permalink]

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New post 27 Jun 2019, 19:46
Thank You AnupamGMAT760 for catering my doubt but there is a constrain question that is :

"No Family Member Will Use Same Car" and there are 5 different CARS for 6 people.

So you have to allocate 1 A brand car to George thus 5!. This is similar to questions, where 3 will come at last position or first position in a 4 digit number and find possible combinations. Please help me understand where I am getting it wrong
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Re: There were 6 members in Mr. George's family: his wife, 2 daughters, on  [#permalink]

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New post 05 Jul 2019, 19:40
Anurag591 wrote:
Thank You AnupamGMAT760 for catering my doubt but there is a constrain question that is :

"No Family Member Will Use Same Car" and there are 5 different CARS for 6 people.

So you have to allocate 1 A brand car to George thus 5!. This is similar to questions, where 3 will come at last position or first position in a 4 digit number and find possible combinations. Please help me understand where I am getting it wrong



Hi Anurag

Consider this:

All family members have a fixed position lets say son gets spot 1, father spot 2 and so on. Now to allocate different cars to them we have to keep these members fixed at their places while we arrange the cars in different spots. Say car A gets spot 1 so it will go to the son since we assigned him that spot and that spot remains fixed. Now there are many ways to approach this:

You can arrange these cars in different spots without constraints in 6!/2! ways. 2! is used since the two A cars are identical and all arrangements involving first A car will be the same as arrangements involving the second A car.

Now factor the constraints, assign Car A to the son so your first spot gets fixed. Now 5 different cars can be arranged in 5 different spots in 5! ways.

We want cases where the son does not get Car A hence (6!/2!) - 5!= 360-120= 240

Alternatively, you also could have used the direct method:

Son A has 4 options to choose from and others still in their respective slots can be assigned a car in 5!/2! ways (5 spots- 5 arrangements- 2 identical)

4*60=240
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Re: There were 6 members in Mr. George's family: his wife, 2 daughters, on   [#permalink] 05 Jul 2019, 19:40
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