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Bunuel
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pristine29
I think it should be :

In first car, 12C4
Second car, 8C4
The remaining four people would automatically get the third car.

For each combination of first car, there are 8C4 combinations of second car (and automatic selection for third).
Therefore, total number of car pooling possibilities will be 12C4 * 8C4.

I too think this is the correct calculation. Can we please get an expert opinion on this pls??
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arunalo
pristine29
I think it should be :

In first car, 12C4
Second car, 8C4
The remaining four people would automatically get the third car.

For each combination of first car, there are 8C4 combinations of second car (and automatic selection for third).
Therefore, total number of car pooling possibilities will be 12C4 * 8C4.

I too think this is the correct calculation. Can we please get an expert opinion on this pls??

I'm not an expert but am certain that the highlighted calculations are correct.

The total number of car-pooling possibilities must be 12c4*8c4*1(for 4c4)
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Bunuel niks18 chetan2u


Quote:
How many different carpool possibilities exist for 12 coworkers from the same neighborhood?

(1) Each of the three cars that will be used for carpooling seats exactly 4 people.
(2) If 2 of the coworkers took the day off, there would be fewer carpool possibilities.

Were this a PS problem would question stem and statement A resulted in 3 * \(12C4\) ?
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adkikani
Bunuel niks18 chetan2u


Quote:
How many different carpool possibilities exist for 12 coworkers from the same neighborhood?

(1) Each of the three cars that will be used for carpooling seats exactly 4 people.
(2) If 2 of the coworkers took the day off, there would be fewer carpool possibilities.

Were this a PS problem would question stem and statement A resulted in 3 * \(12C4\) ?

This basically asks to distribute 12 in 3 groups of 4 each..
First group-12C4
Second - 8C4.. 4 out of remaining 8
Third remaining 4
So 12C4*8C4..

Direct formula for these (ax) items to be distributed in a groups of x items each is (ax)!/(x!)^a... Here (3*4)!/(4!)^3

12C4*8C4=12!/(8!4!) * 8!/(4!4!)=12!/(4!)^3..

You will find this in my signature about combinations
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adkikani
Bunuel niks18 chetan2u


Quote:
How many different carpool possibilities exist for 12 coworkers from the same neighborhood?

(1) Each of the three cars that will be used for carpooling seats exactly 4 people.
(2) If 2 of the coworkers took the day off, there would be fewer carpool possibilities.

Were this a PS problem would question stem and statement A resulted in 3 * \(12C4\) ?

This basically asks to distribute 12 in 3 groups of 4 each..
First group-12C4
Second - 8C4.. 4 out of remaining 8

Third remaining 4
So 12C4*8C4..

Direct formula for these (ax) items to be distributed in a groups of x items each is (ax)!/(x!)^a... Here (3*4)!/(4!)^3

12C4*8C4=12!/(8!4!) * 8!/(4!4!)=12!/(4!)^3..

You will find this in my signature about combinations




why are we not arranging them further?
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