Question

The La Suerte family has 13 members. They plan to use their 13-seater van during a trip. Family members who know how to drive must occupy the front seats. Children should occupy the next two rows while the other members should be seated at the back. In how many ways will the family members be seated if only three of them know how to drive and 6 of them are children? The van has 4 rows with three seats each row for the first three rows, and 4 seats at the back row.

Answer 1

"The van has 4 rows with three seats each row for the first three rows"
in front: 3 seats per row

"only three of them know how to drive"
so, the 3 that know how to drive are always in the front row
also nobody else is there see above

For the first row, there are 3 people and 3 seats, they are always
the same but they can change seats in some combinations.

Answer 2

@mathessentials I still don't understand it...

Answer 3

OK do you understand what is in my post though?

Answer 4

@mathessentials Yeah.

Answer 5

so we have to get this to the form of choosing marbles from a pot.... because that's the form combinatorics work with.

I guess we can say "first pick" will be the same like "person one gets"

in the marble and pot problem there are different possible scenarios which one is picked "first". similarly, there are different scenarios which seat will be used by "person one".
In the marble problem, there are 3 picks. Here there are also 3 picks for the 3 seats, just we don't care about "first, second, third" we care about "driver1, driver2, driver3" but it is the same structure for combinatorics

Answer 6

In the marble problem, first pick can have marble1,marble2,marble3

here, the driver1 can get seat1, seat2, seat3

same amount of combinations/possible ways

Answer 7

The front row is a 3 choose 3 scenario

are you familiar with "choose" in combinatorics?

Answer 8

@mathessentials Yes, it should be 3C3?

Answer 9

yes. it's also without replacement.
a seat is taken and no longer available when it is being chosen.

Answer 10

@mathessentials What's next?

Answer 11

Next, you do similar reasoning for the remaining rows in the van.

also, the # of ways of each row are finally multiplied together

for 1,2,3 and A, B

1,2,3 - A
1,3,2 - A
2,1,3 -A
2,3,1 -A
3,1,2 -A
3,2,1 -A

1,2,3 -B
1,3,2 -B
... just as many again, so whatever combinations for 1,2,3
times the combinations of A,B

in this example

Answer 12

sorry, 3C3 is wrong

in n choose k, order doesn't matter
for us order matters

since I said first pick <=> first driver
second pick <=> second driver etc.

it matters in which order and so it must be something else

Answer 13

must be a permutation instead because position of the drivers matters

Answer 14

ok? :)

Answer 15

\[permutations:\frac{ n! }{ \left( n-r \right)! }\]with no repetition
with order matters

n is the number of things to choose from (seats in front row =3)
r is the number of things you choose (3 seats for 3 drivers, =3)

Answer 16

^ this is for the first row part

then there's the rows 2,3 for the children

In that part of the problem there are: 6 seats to choose from
how often? for 6 children to go for each seat

!! it is a permutation though, NOT 6C6 !!

Answer 17

@mathessentials So its 6P6?

Answer 18

Yes :)

6P6 with "no repetiton"
for 2nd part of the problem

Answer 19

3P3 for first part of the problem

Answer 20

--Family members who know how to drive must occupy the front seats. Children should occupy the next two rows

"three of them know how to drive"
first row all drivers

"and 6 of them are children"
rows 2+3 all children

no child allowed in row 4
illegal permutation

Answer 21

so that just leaves 3P3 and 6P6, since there is no empty seat
and hard requirements

Answer 22

... for the drivers and children.

Then we also need the remaining passengers.

Answer 23

@mathessentials What is the final equation for the problem?

Answer 24

well the third part is the remaining passengers and remaining seats

remaining passengers: 13-3 drivers - 6children
= 13 -9 = 4 other
and there's 13-seater -3-6 = 4 seats

I think everyone not child or driver should be in back row:
4P4

Answer 25

and then we need to combine the 3P3, 6P6 and 4P4

Answer 26

@mathessentials Add them?

Answer 27

I think we should multiply them because we care about the total seating in the van, like, if children and passengers remained the same but two drivers change position, the whole situation was changed

the seating should be a sequence of driver seating, children, passengers
they depend on each other to be unique

Answer 28

"In how many ways will the family members be seated "

Answer 29

if you add them you basically get the options a person has
the # of different seats he can get

Answer 30

@mathessentials But if we multiply them the answer is 103680..

Answer 31

simple example:
´´´´´´´´´´´´´´
front: 1,2,3
back: A, B

combinations front: 1,2,3&1,3,2&... (6 combinations)
combinations back: A,B & B,A (2 combinations)

combined unique different scenarios:
6 times A,B and
6 times B,A

if we added them
1,2,3 A, B
1,3,2 B, A
2,1,3
2,3,1
3,1,2
3,2,1

they wouldn't have much to do with each other
they would be separate, isolated choices

Answer 32

I think you should only add if it is different cases like

you can go to bakery and choose from 3 breads
or
go to supermarket and choose from 2 toast

total possible choices: 5 total choices
the different 3/2 have nothing to do with each other

Answer 33

but they are both possible so add them

Answer 34

new example