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Family Seating

by tritrantran » Sun Dec 14, 2008 3:47 pm
What's a good way to solve these type of problems?

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?

A) 28
B) 32
C) 48
D) 60
E) 120




OA 32
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by Neo2000 » Sun Dec 14, 2008 6:00 pm
Since either one of the parents must drive, a parent can be selected in [spoiler]2C1 = 2ways[/spoiler]

Since the 2 daughters refuse to sit together, figure out the separate ways the daughters do NOT sit together.

Start with the back-seat:
Both daughters in the back seat(near each window) can be arranged in 2ways.
The middle place can be filled[spoiler] either by the Son or the other parent. This can be done in 2C1 = 2ways.
So total no: of ways such that both daughters sit in the back = 2x2x2 = 8
[/spoiler]
Front Seat:
Pick 1 daughter to sit in the front seat:[spoiler] 2C1 = 2[/spoiler]
The parent, son and daughter can now be arranged [spoiler]in the back seat in 3! ways.
Total no: of ways such that 1 daughter sits in the front and the other daughter sits in the back = 2x2x6 = 24

Therefore, total number of seating arrangements = 24+8 = 32[/spoiler]
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Family Seating

by ifairo » Wed Dec 17, 2008 8:47 pm
The driver seat can be filled in 2 ways (Mother or Father). The remaining 4 seats can be filled in 4! ways. Total arrangement is 2 * 4! = 48 ways.

We are not done yet.

We need to eliminate the number of arrangements in which the two daughters sit next to each other in the back seat.

After we filled the driver seat, we have the navigator seat, and the remaining 3 back seats vacant. The navigator seat can be filled by either one of the parent and by the son, i.e., 2 * 1 = 2 ways.

The front seat arrangement comes to 2 * 2 = 4 ways.

Assuming the daughters sit next to each other, can be considered as "1" instead of 2, we have 2 consolidated elements to jumble. That leads to 2! ways, and the two daughters collapsed to one can be arranged in 2! ways - D1D2 or D2D1.

So, there are 2! * 2! ways to arrange the back seat with two daughters sitting next to each other.

Overall, there 4 * 4 = 16 ways in which the two daughters sit next to each other.

Remove this from the actual 48 ways. So the answer is (B) 32.
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by abcdefg » Fri Jul 10, 2009 6:39 am
Neo, you logic seems to be right.

- For the condition to be met, a) either the 2 daughters must be sitting in the back both next to a window or b) 1 of the daughters must be sitting in the front. So let's add up the number of ways that a) and b) can happen.

a) can happen like you said if the daughters sit next to the window (2 ways) and 1 parent and son fill in the middle (2 ways). 2x2 is 4.
b) can happen if we have 1 daughter sit in the front (2 ways) and daughter son and parent sit in the back (3! arrangements). 2x3! = 2x3x2 = 12 ways.

Add up a) and b) we get 4+12 = 16.

The last thing we need to consider is the number of ways to choose a driver, 2. 2x16 = 32.

Please tell me if my logic makes sense. This was a very messy problem.
Neo2000 wrote:Since either one of the parents must drive, a parent can be selected in [spoiler]2C1 = 2ways[/spoiler]

Since the 2 daughters refuse to sit together, figure out the separate ways the daughters do NOT sit together.

Start with the back-seat:
Both daughters in the back seat(near each window) can be arranged in 2ways.
The middle place can be filled[spoiler] either by the Son or the other parent. This can be done in 2C1 = 2ways.
So total no: of ways such that both daughters sit in the back = 2x2x2 = 8
[/spoiler]
Front Seat:
Pick 1 daughter to sit in the front seat:[spoiler] 2C1 = 2[/spoiler]
The parent, son and daughter can now be arranged [spoiler]in the back seat in 3! ways.
Total no: of ways such that 1 daughter sits in the front and the other daughter sits in the back = 2x2x6 = 24

Therefore, total number of seating arrangements = 24+8 = 32[/spoiler]
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Re: Family Seating

by doclkk » Fri Jul 10, 2009 3:30 pm
tritrantran wrote:What's a good way to solve these type of problems?

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?

A) 28
B) 32
C) 48
D) 60
E) 120




OA 32
Can anyone do this via MGMAT anagram method?
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Re: Family Seating

by doclkk » Fri Jul 10, 2009 3:43 pm
ifairo wrote:The driver seat can be filled in 2 ways (Mother or Father). The remaining 4 seats can be filled in 4! ways. Total arrangement is 2 * 4! = 48 ways.

We are not done yet.

We need to eliminate the number of arrangements in which the two daughters sit next to each other in the back seat.

After we filled the driver seat, we have the navigator seat, and the remaining 3 back seats vacant. The navigator seat can be filled by either one of the parent and by the son, i.e., 2 * 1 = 2 ways.

The front seat arrangement comes to 2 * 2 = 4 ways.

Assuming the daughters sit next to each other, can be considered as "1" instead of 2, we have 2 consolidated elements to jumble. That leads to 2! ways, and the two daughters collapsed to one can be arranged in 2! ways - D1D2 or D2D1.

So, there are 2! * 2! ways to arrange the back seat with two daughters sitting next to each other.

Overall, there 4 * 4 = 16 ways in which the two daughters sit next to each other.

Remove this from the actual 48 ways. So the answer is (B) 32.
OK - front seat = 4 ways. - but why is the front seat a restriction at all.

So is the back seat 3C2?

Can you explain this again. I got to 48 and I'm lost.
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by shibal » Sat Jul 11, 2009 6:13 am
when should I multiply or add the answers in this situation?
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by gmat740 » Sat Jul 11, 2009 9:15 am
when should I multiply or add the answers in this situation?
When considering two independent situations,you must add.

Trick : Ask yourself the question :
this way is possibe or(+) this way is also possible or(+)....

I hope I have explained my point.
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by paras747 » Sun Nov 06, 2011 10:47 am
The easiest way to solve this question is to consider the restrictions separately. Let's start by considering the restriction that one of the parents must drive, temporarily ignoring the restriction that the two sisters won't sit next to each other.

This means that...
2 people (mother or father) could sit in the driver's seat
4 people (remaining parent or one of the children) could sit in the front passenger seat
3 people could sit in the first back seat
2 people could sit in the second back seat
1 person could sit in the remaining back seat

The total number of possible seating arrangements would be the product of these various possibilities: 2 × 4 × 3 × 2 × 1 = 48

We must subtract from these 48 possible seating arrangements the number of seating arrangements in which the daughters are sitting together. The only way for the daughters to sit next to each other is if they are both sitting in the back.

This means that...
2 people (mother or father) could sit in the driver's seat
2 people (remaining parent or son) could sit in the front passenger seat

Now for the back three seats we will do something a little different. The back three seats must contain the two daughters and the remaining person (son or parent). To find out the number of arrangements in which the daughters are sitting adjacent, let's consider the two daughters as one unit. The remaining person (son or parent) is the other unit. Now, instead of three seats to fill, we only have two "seats," or units, to fill.
There are 2 × 1 = 2 ways to seat these two units.
However, the daughter-daughter unit could be d1d2 or d2d1
We must consider both of these possibilities so we multiply the 2 by 2! for a total of 4 seating possibilities in the back.
We could also have manually counted these possibilities:
d1d2X, d2d1X, Xd1d2, Xd2d1

Now we must multiply these 4 back seat scenarios by the front seat scenarios we calculated earlier:
(2 × 2) × 4 = 16
front back

If we subtract these 16 "daughters-sitting-adjacent" scenarios from the total number of "parent-driving" scenarios, we get: 48 - 16 = 32

The correct answer is B.
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by GMATGuruNY » Mon Nov 07, 2011 3:51 am
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