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35%
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Question Stats:
67% (00:58) correct
33%
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The Coen family consists of a father, a mother, two children and a dog. A photographer is about to take the family's picture. How many different arrangements (of standing in a row) does the photographer have, if it is known that the father insists of standing by his woman, as the song says?
A. 12
B. 24
C. 48
D. 60
E. 120
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A. 12
B. 24
C. 48
D. 60
E. 120
_________________
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Bunuel
We need to find the number of arrangements of Father, Mother, Child-1, Child-2, and a Dog, so that F and M are always together.
Consider F and M to be one unit, then we would have total of four units: {FM}, {C1}, {C2}, {D}. The number of arrangements of these four units is 4!, but F and M within their unit can be arranged in 2 ways: {FM} or {MF}, so total number of arrangements is 4!*2=48.
Answer: C.
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General Discussion
Posts: 13
F, M, C1, C2, D (let those be Father, Mother, Children 1, Children 2 and Dog respectively)
We need to find all the variations for:
F, M, X, X, X --> the Arrangement for the three spots is 3!, so 6 options. Also need to consider the other 6 options when mother is at position one and father at 2. So in total 12.
X, F, M, X, X --> the same
X, X, F, M, X --> the same
X, X, X, F, M --> the same
The total number of variatons would be 12*4, which is 48.
We need to find all the variations for:
F, M, X, X, X --> the Arrangement for the three spots is 3!, so 6 options. Also need to consider the other 6 options when mother is at position one and father at 2. So in total 12.
X, F, M, X, X --> the same
X, X, F, M, X --> the same
X, X, X, F, M --> the same
The total number of variatons would be 12*4, which is 48.
GMAT 1: 760 Q51 V42
Posts: 10,057
Total people: \(F, M, C_1, C_2, D\) [Row arrangement]
Condition: Father insists on standing by his woman (F, M) forms 1 group
=> C_1, C_2, D, (F, M) = arranging 4 people will take 4! ways and F, M can be arranged in 2! ways.
Total ways: 4! * 2! = 24 * 2 = 48
Answer C
Condition: Father insists on standing by his woman (F, M) forms 1 group
=> C_1, C_2, D, (F, M) = arranging 4 people will take 4! ways and F, M can be arranged in 2! ways.
Total ways: 4! * 2! = 24 * 2 = 48
Answer C
Iector
Joined: 10 Jul 2020
Last visit: 06 Jul 2022
Posts: 1
Location: Russian Federation
Schools: HEC MiM "24
Schools: HEC MiM "24
Posts: 1
Family has five members. Two members (mother&father) must be next to each other. So, we should consider them as one member, who can be arranged in two variations (mother&father; father&mother). Now we have four members and four spots for them __ __ __ __. The first spot can take 4 members, the second - 3, the third - 2, and the last - 1. It will be 4*3*2*1=4!=24 arrangments, but we should remember that our "unique" member (which includes mother&father) can be in two variations, so 24*2=48.
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