Combination Manhattan CAT #5 problem 31

[Guest]

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?
28
32
48
60
120

Could you please explain this solution. The current explanation does not consider the possible if one of the daughters were to sit in the front seat.

[RonPurewal]

for the benefit of everyone reading the forum, i've posted the solution from the practice exam website (verbatim) below.

note the way this solution proceeds: it considers all the ways in which the two daughters can sit together, and then subtracts those possibilities from the total number (because those are, after all, the possibilities that we don't want).

this is why we don't explicitly have to think about one of the daughters in the front seat: if that happens, then the two daughters can't sit together. since we're figuring out the ways the daughters can set together (so that we can eventually subtract them), we don't need to enumerate those possibilities.

-- original solution follows --

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.

[Guest]

Is there a way to solve this question using the anagram grid? (e.g., YYNN)

[RonPurewal]

Is there a way to solve this question using the anagram grid? (e.g., YYNN)

you can't do it with the straight anagram grid, no, because there are restrictions on the choices. the anagram grid, in pure form, only works when the choices are completely unrestricted (i.e., anyone can go anywhere).

it's possible that you could divide the problem into enough separate cases that you could use an anagram grid for each case, but that would be a lot of cases indeed.

[NewSc2]

Hi, sorry to bump this, but I'm stuck at the end of this problem.

My initial reasoning:
1) Father/Mother have to sit in front, so treat them as one unit (for now). Therefore 4! = 24, and multiply by 2 to account for one with father at head, one with mother. So 48 total possibilities with Father/Mother in front.

2) [initial thought]: So now to subtract the possibilities of sisters together:
Glue the sisters together, so among the 4 leftover possibilities, there are 3! = 6. Multiply by 2 (sister arrangement) to get 12. If there had been an answer choice of 36, I would've picked it and moved on. Alas, there wasn't, but I was pretty confident in the 48 figure, so I guessed between answer choices 28 and 32 and got this question right.

But now my reasoning, after reading the explanation:
2) If one of the sisters is in the front seat, they will not be sitting next to each other. So among the 3 back seats, there are 4 possibilities (S1, S2, other / S2, S1, other and other, S1, S2 / other S2, S1). So I understood the explanation up to here.

But why do you multiply this figure (4) by 2 x 2? The explanation cites a "Now we must multiply these 4 back seat scenarios by the front seat scenarios we calculated earlier ... 2 people (mother father in driver seat) ... 2 people (remaining parent or son) in the passenger" but aren't there more possibilities? e.g., Mom sits in "other" seat, Dad sits in "other" seat, and Son sits in "other" seat?

Could you elaborate on this further?

[mschwrtz]

If I correctly understood your reasoning, you have already assigned a parent to the front seat (remember that you multiplied by 2 earlier in the solution to reflect that either parent might be there?). You only have one parent left now.

[mschwrtz]

Oh, and just to verify by brute (or brutish) force:

With F driving, MS1S2, BS1S2, S1S2M, S1S2B.
With M driving, FS1S2, BS1S2, S1S2F, S1S2B.

Same as 2--because two different parents--times PS1S2, BS1S2, S1S2M, S1S2B.

[avihil]

My count is 40.
I'll explain

the # of options is blocked by 48 as written (2 drivers + 4!)

Now for the other 4 seats (Max options is 24)
Lets subtract the limited sitting permutation:

2 sisters one next to another in the back seat.

This is 2 options multiple by 2! (of replacing the sisters in place)
so it's (24-4)2 = 40
??

Where am i wrong ?

Thx



----

OK - understood :-)
==> Every option for the sisters to sit together should be multiple by 2 ( of the brother and the not driving parent )

[jnelson0612]

Great, you figured it out and answered your own question! That is always the best way to learn.

[ayush.rastogi.13]

Hi,

Below is my method to solve this question.

Let Father = F, Mother = M, Daughter 1 = D1, Daughter 2 = D2 & Son = S.

Case 1 :
M&F occupy front seats. Therefore 2*1 = 2 (since back seats can be occupied in one way)
Case 2 :
Either M or F take the driving seats and the other front seat by son. Therefore 2*1*1 = 2
Case 3 :
Either M or F take the driving seat and the other front seat by one of the daughters. Therefore 2*2*3! ( since anyone can sit at the 3 back seats)
= 2*2*6 = 24

Adding all cases my answer is 2+2+24 = 28.

Please help me find out the mistake I made. Or may be I have missed another case. Thanks in advance !!

[RonPurewal]

i'm going to rename the daughters "sarah" and "leila", since those names are so much prettier than "d1" and "d2".
and let's let the son be "mark".

Case 1 :
M&F occupy front seats. Therefore 2*1 = 2 (since back seats can be occupied in one way)

in each of these cases, the back seats can be occupied in two ways, not one.
from left to right, you can have...
... sarah, mark, leila
or
... leila, mark, sarah
these are clearly not the same. so this case accounts for 4 possibilities, not 2.

Case 2 :
Either M or F take the driving seats and the other front seat by son. Therefore 2*1*1 = 2

same problem as case 1, so this is also supposed to be 4 outcomes.

case 3 is fine. so, there are actually 4 + 4 + 24 = 32 possibilities.

[ayush.rastogi.13]

ohh Yes !! Thanks for the explanation.

[jlucero]

Glad it helped.

[DouglasM819]

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

Hello,

Can someone tell me whether this logic would always work or just happened to work for this problem.

In the books, in a less complicated example there it has 6 people in 6 seats but 2 people don't want to sit next to each other. So you have 6! with one constraint so we have 6! - (2*5!) 480. Aka this is lumping the two people as one and subtracting the problem solutions.

Given that, I addressed this problem as:

Driver: 2
Passenger: 4
Back: 3*2*1
So (2*4)*(3!) = 48 before accounting for problem outcomes

Instead of altering the first two seats, I focused only on problem solutions in the back seat using the similar approach outlined in the book. I wrote 3!-2! = 6-2 =4, then 8*4 =32. I realize that this was prob a mistake as this should of produced 3!-2(2!) = 2, thus 8*2 =16.

In this case it worked obviously.
My questions are
1.) is 2*4*(3!-2!) fundamentally setup wrong?
2.) Did I get this problem right by accident because the numbers just happened to work?
3.) Forward thinking to other problems, would a set up like this where I left the 2*4 untouched and focused only on the back seat work with a slight fix?

I hope all this makes sense.

Thanks.

[RonPurewal]

1.) is 2*4*(3!-2!) fundamentally setup wrong?

yep. lucky.

if you want to go this route, you can't use the "4", because the numbers are different if the person in the passenger seat is one of the girls.

so, break it up into two cases.

• if the passenger seat is not occupied by one of the girls:
2 • 2 • (3! – 2(2!))
you need 2(2!), not just 2!, because you can switch the two girls' spots. but you seem to understand that already.
this is just 2•2•2 = 8.

• if the passenger seat is occupied by one of the girls:
2 • 2 • 3!
you don't have to exclude any cases, since there's no hate in the back seat anymore.
this is 24.

8 + 24 = 32.