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taking a road trip in a sedan

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by sanju09 » Sat Aug 14, 2010 2:19 am
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


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by shibal » Sat Aug 14, 2010 7:35 am
this is a gmatprep question btw...

i found it easier drawing it:

M F
DSD (has 2 arrangements)

M S
DFD (has 2 arrangementes)

M D
DFS (3C2 * 2 (times 2 bc there are two different daughter - has 12 arrangementes)

F M
DSD (has 2 arrangements)

F S
DMD (has 2 arrangementes)

F D
DMS (3C2 * 2 (times 2 bc there are two different daughter - has 12 arrangementes)

add everything up, u'll have 32 different arrangements... hope it helps
best
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by Prashantbhardwaj » Sat Aug 14, 2010 9:55 am
There are 2 arrangements possible

I) parents sit up front:

[2 * 1 * 2*1*1] * 2 = 8 arrangements

Front Seats(Objects for consideration: 2 parents, parent left)

Rear seats (objects for consideration: 2 daughters, son, daughters left i.e. 1)

All multiplied by 2 because the 2 daughters are diff.

II) parents sit separate:

[2 * 2 * 3*2 * 1] * 2 = 24 arrangments

Fornt seats (objects for consideration: 2 parents, 2 daughters )

Rear seats (Objects for consideration 3 persons: mother, daughter and son)[/u]

All multiplied by 2 for similar reasons as above.

Hence total 32
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