family trip ..

[aru4912]

A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?


1.28
2.32
3.48
4.60
5.120

Friends, pl help

[lsr]

A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?

1.28
2.32
3.48
4.60
5.120

Friends, pl help

There are 32 different arrangements.

[KBTA]

IMO
32

When we taking one daughet in the fron seat
Total ways= 2*2*3!=24

Other case
Total ways=2*2*3!-2*2*2!*2!=8

So, finally ways = 24+8=32

[Name User]

A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?


1.28
2.32
3.48
4.60
5.120

Friends, pl help

Case 1:
Boy back seat:
Mom Drive: 2 ways
Dad drive:2 ways

Case 2:
Boy font seat
mom drive: 2 ways
Dad drive:2 ways

Case 3:
Girl#1 front seat
Mom drive: 6 ways
Dad drivEL 6 ways

Case 4:
girl #2 front seat
Same as above
12+12 <==from case 3&4
8 <==From case 1,2

Total: 24+8=32 ways

[LegoLife]

My answer is 32.

The number of arrangements will be like: 2*4*2*2*1 after all restrictions.

Hence, 32.

lsr, please let me know your approach to this problem.I'd appreciate any shorter method.

[lsr]

My answer is 32.

The number of arrangements will be like: 2*4*2*2*1 after all restrictions.

Hence, 32.

lsr, please let me know your approach to this problem.I'd appreciate any shorter method.

Number of ways the two daughters will sit together.

(2)*(2)*(2!)*(2!) = 16

2 for the driver-seat.
2 for the passenger-seat.
3 for the back seat with 2 of them sitting together: (3-2+1)! = 2!
2! ways for the 2 daughters to arrange themselves.

Total number of ways for the family to sit = 2*4! = 48 (since only two can sit in the driver-seat).

Number of ways the family can sit without the two daughters sitting next to each other = 48-16 = 32.

Lego, I find that with restrictions, many times it is faster to count the number of arrengments of those restrictions and subtract from the total, than it is to add all the individual cases that would comply with the restricitions.

[LegoLife]

Thanks for the wonderful explanation lsr !

I agree that in restrictions, its easier to find the total first and then subtract the part with restrictions.The same goes for Probability too : 1 - p(n).

[nikiforos]

Another vote for 32