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Stockmoose16
- Master | Next Rank: 500 Posts
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Here's a very difficult combinatrics problem from an MGMAT CAT:
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?
I tried to solve it by figuring out the number of ways the people could be arranged. Here's my logic, please explain why it's wrong:
Driver's Seat: 2 people (Mother or Father)
Passenger Seat: 4 people (M/F (whoever not picked above), plus 3 kids could sit in passenger seat)
Back Left seat: 2 people (assuming mother got picked to sit in passenger seat, one of the sisters must sit in back end seat so he/she doesn't sit next to other sister)
Back Middle Seat: 1 person (since the girls can't sit together, brother must sit in middle)
Back Right Seat: 1 person (last remaining sister)
So: 2*4*2*1*1 = 16. Since the sisters are interchangeable, you need to multiply by the number of ways they can be arranged, 2!. So the answer is 32 (which is correct). However, something seems to be wrong with this logic. How would you account for the situation where one of the sisters is selected for the passenger seat, and then any of the remaining three can be arranged in the back seat in any order?
The answer key says to tie the two sisters together as one unit, where you'd find all the ways the two of them can sit together (which can only be in the back seat). You'd then subtract from the number of ways the girls can sit together in the back seat from the total number of arrangements.
What's wrong with my method?
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?
I tried to solve it by figuring out the number of ways the people could be arranged. Here's my logic, please explain why it's wrong:
Driver's Seat: 2 people (Mother or Father)
Passenger Seat: 4 people (M/F (whoever not picked above), plus 3 kids could sit in passenger seat)
Back Left seat: 2 people (assuming mother got picked to sit in passenger seat, one of the sisters must sit in back end seat so he/she doesn't sit next to other sister)
Back Middle Seat: 1 person (since the girls can't sit together, brother must sit in middle)
Back Right Seat: 1 person (last remaining sister)
So: 2*4*2*1*1 = 16. Since the sisters are interchangeable, you need to multiply by the number of ways they can be arranged, 2!. So the answer is 32 (which is correct). However, something seems to be wrong with this logic. How would you account for the situation where one of the sisters is selected for the passenger seat, and then any of the remaining three can be arranged in the back seat in any order?
The answer key says to tie the two sisters together as one unit, where you'd find all the ways the two of them can sit together (which can only be in the back seat). You'd then subtract from the number of ways the girls can sit together in the back seat from the total number of arrangements.
What's wrong with my method?
