A family with 1 mother, 1 father, and 3 children sit in a car with 2

Question

A family with 1 mother, 1 father, and 3 children sit in a car with 2 front seats and 3 back seats. If the mother drives with a child in the front, and the father sits in the back with the other two children, how many seating arrangements are possible?

(A) 3
(B) 4
(C) 6
(D) 12
(E) 18

(A) 3
(B) 4
(C) 6
(D) 12
(E) 18

fantaisie · Answer

I wasn't sure if the order of the way children are arranged matters in this question. Since children were mentioned as one (imagine we have 1 RED marble, 1 BLUE and 3 GREEN) and the front two seats are already determined, then there are 3 ways how the father can be seated in the back seat with regards to his children:

F C C
C F C
C C F

But if the order matters, then my solution doesn't hold. I am looking forward to discover the actual solution!

RafaelAraujo10 · Answer

In the front seat we have 3 possibilities of children. (C1 or C2 or C3)
In the back seat we have (2!*1)*3 = 6 possibilities ( C1,C2,F / C2,C1,F / C1,F,C2 / C2,F,C1 / F,C1,C2 / F,C2,C1 -> IF C3 is in front seat)
So, 6*3 = 18 (try to figure out to change the possibilities of sitting back the C3 child in the place of C1 or C2 and do the 6 possibilities again and again).

Or, another way to see:
Looking into the back seat:
( C1,C2,F / C2,C1,F / C1,F,C2 / C2,F,C1 / F,C1,C2 / F,C2,C1) = 6
or +
( C3,C2,F / C2,C3,F / C3,F,C2 / C2,F,C3 / F,C3,C2 / F,C2,C3) = 6
or +
( C1,C3,F / C3,C1,F / C1,F,C3 / C3,F,C1 / F,C1,C3 / F,C3,C1) = 6
-------
18
Letter E :-D

glatzekatze · Answer

Seat of mother is fixed
one child is in the front but we don't know which one.
So lets consider child one is in the front, so we have 3! = 6 arrangements in the backseat
then consider child two is in the front, again we have 3! = 6 arrangements
And the same for the last one. 6*3 = 18, hence E

chetan2u · Answer

Hi,

Your answer would have been correct had we been looking at identical things, the way you have mentioned about RED marble..
But here the order of chidren will also matter as these are different identities..
so 3*3!... 3! is the way these 3 kids can be arranged within themselves

rocko911 · Answer

If the question asks about number of arrangements then why are we not considering the mothers seat combination ?

We got 18 but there are 3 combinations of either of 3 child sitting with the mother, Then we the answer is not 21?

ScottTargetTestPrep · Answer

We can let F be the father, M be the mother, and A, B, and C be the 3 children. We are given that the mother must occupy the driver’s seat and one of the children is sitting in the front. 
 
So, let’s assume that M and A are sitting in the front; this leaves F, B, and C sitting in the back. The front row seats are fixed since M must be at the driver’s seat, but the back row seats are not. If F, B, and C are sitting in the back, there are 3! = 6 ways to occupy the back row seats. 
 
Similarly, if M and B are sitting in the front and F, A, and C are sitting in the back, there will be also 6 ways to occupy the back row seats. Lastly, if M and C are sitting in the front and F, A, and B are sitting in the back, there will be another 6 ways to occupy the back row seats.
 
Thus the total number of seating arrangements is 6 + 6 + 6 = 18. 
 
Answer: E

HAPPYatHARVARD · Answer

In the back seat we can pic two children in 3C2 ways and F C C can be arranged in 3! ways.
So, 3C2 x 3! ways.
Since the child can not drive so, C M can be arranged in one one way.
so ans is 3C2 X 3!=18

bumpbot · Answer

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A family with 1 mother, 1 father, and 3 children sit in a car with 2

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