Five siblings Abe, Beth, Carl, Dave and Enigma are to be seated in a

Question

Five siblings Abe, Beth, Carl, Dave and Enigma are to be seated in a row. If Abe and Beth are to be seated together and Carl and Dave cannot sit together, in how many ways can the siblings be seated in a row?

A. 12
B. 18
C. 24
D. 36
E. 48

A. 12
B. 18
C. 24
D. 36
E. 48

IanStewart · Accepted Answer

First, if A and B are going to sit together, we can consider them as a single letter, AB. Then we have 4 letters, and 4! = 24 arrangements in total. But they could also sit in the order BA, so we have another 24 orders, for 48 in total.

Now if A and B will sit together, and C and D will also sit together, we are just arranging three letters, AB, CD and E, so we'd have 3! = 6 arrangements in total, but we have to multiply by 2 twice, once because A and B could sit in two orders, and once because C and D can sit in two orders, so there are 24 arrangements where A and B are together and C and D are together. So, subtracting from the 48 we got in the first step, there must then be 48 - 24 = 24 arrangements where A and B are together and C and D are not together.

nick1816 · Answer

Number of ways when A and B sit together= 4!*2!=48
Number of ways when A and B sit together and when C and D sit together= 3!*2!*2!= 24
Number of ways when A and B sit together but when C and D do not sit together= 48-24=24

GMATWhizTeam · Answer

Solution:

Let us assume the 5 siblings be  A, B, C, D  and  E .

We are given that  A  and  B  are to be seated together. So we can assume  AB  to be a single entity. Because of which there will essentially be 4 entities:  AB, C, D  and  E

So the number of ways in which these  4  entities can be arranged  = 4!	imes 2!=48

Now there is an additional condition that Carl and Dave or  C  and  D  cannot sit together. So let us make them sit together and calculate the unfavorable cases.

So we can assume  CD  to be a single entity. Because of which there will essentially be 3 entities:  AB, CD  and  E .

So number of ways in which these  3  entities can be arranged  = 3!	imes 2!	imes 2!=24

Thus the  number of favorable cases = total cases - unfavorable cases

⇒ 48-24

⇒ 24

Hence the right answer is  Option C .

PSKhore · Answer

A B _ _ _
_ A B _ _
_ _ A B _
_ _ _ A B

AB have 4 positions and they can rearrange themselves in 2 ways. Therefore AB can be seated in 4*2 = 8 ways
Now, since C & D cannot be seated together, E will occupy the place between them. 
In each of the above cases C/D will have 3 positions to take and the remaining two will adjust themselves.
Total ways = 8*3 
= 24

mvgkannan · Answer

@  IanStewart  @  GMATWhizTeam  I have a doubt regarding the wording of the problem. Since the problem says, If Abe and Beth are to be seated together and Carl and Dave cannot sit together, I believe it is only one of the conditions, the other unstated one goes that if A and B are not seated together, C and D can be together or not! That means the answer will be 96 if I am not wrong and. Since it is a row, Left/Right shouldn't matter, then answer would be 96/2 = 48.

No of ways all siblings be seated without restriction = 120

No of ways all siblings be seated violating the restriction, i.e, A & B together and also C and D also together is 24.
Total favourable cases = 96.
Since it is a row, Left and right shouldn't matter, hence answer will be 48. 
Please correct me if I am wrong and where I am wrong.

Five siblings Abe, Beth, Carl, Dave and Enigma are to be seated in a

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Five siblings Abe, Beth, Carl, Dave and Enigma are to be seated in a

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