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In how many ways can Ann, Bob, Chuck, Don and Ed be seated

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In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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New post 25 Nov 2010, 19:06
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In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

A. 24
B. 48
C. 56
D. 72
E. 96

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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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New post 25 Nov 2010, 21:55
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shrive555 wrote:
In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

24
48
56
72
96


When the constraint on an arrangement says, "Two people should not be seated together," we do just the opposite. We make them sit together! Kind of tie them with a thread and assume they are one unit!
Let's see why....

These 5 people can be arranged in 5! ways. These are the total number of ways you get.
Now, when we tie 2 people together, we have only 4 entities to arrange. We can do this in 4! ways. But in each of these entities, the two people can sit in two different ways (AB and BA). So number of ways in which these two people sit together is 4!*2!.

Now, the ways in which these two people will not be together will be 5!- 4!*2! = 4!(5 - 2) = 72
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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New post 26 Nov 2010, 13:22
my solution:

3! * 6*2 = 72

3! for Chuck, Don and Ed

6 positions for Ann and Bob: AC AD AE BC BE CE * 2 (as they can switch they positions)

It is not as concise as Karishma's but it works.
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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New post 27 Nov 2010, 14:13
craky wrote:
my solution:

3! * 6*2 = 72

3! for Chuck, Don and Ed

6 positions for Ann and Bob: AC AD AE BC BE CE * 2 (as they can switch they positions)

It is not as concise as Karishma's but it works.


Yes, your logic is absolutely fine. If the number of people is manageable, we can directly find the number of ways in which the two of them should not be seated together.
You arranged C, D and E in 3! ways.
Attachment:
Ques2.jpg
Ques2.jpg [ 3.56 KiB | Viewed 5944 times ]


The 4 dots show 4 positions for Ann.When Ann occupies one of these positions, 3 positions are left over for Bob. So Ann and Bob can sit in 4*3 = 12 ways.
Total 3! * 12 = 72 ways.
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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New post 13 Oct 2015, 20:24
tuanquang269 wrote:
In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

(A) 24

(B) 58

(C) 56

(D) 72

(E) 96


In these types of questions, always try to subtract the not allowed arrangements from the total arrangements. This makes our lives much easier.
Here we will assume that Ann and Bob always sit together i.e. they are a single entity
Total arrangements: 5!

Arrangements in which Ann and Bob sit together = 4!*2!
We get this because, total arrangements =4! and Ann and Bob can be arranged in 2! ways

Subtracting this from the total: 5! - 4!*2! = 72. Option D
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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New post 19 Apr 2017, 16:02
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shrive555 wrote:
In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

A. 24
B. 48
C. 56
D. 72
E. 96


We can use the following formula:

Total number of ways to arrange the 5 people = (number of arrangements when Bob sits next to Ann) + (number of arrangements when Bob does not sit next to Ann)

Let’s determine the number of arrangements when Bob sits next to Ann.

We can denote Ann, Bob, Chuck, Don, and Ed as A, B, C, D, and E, respectively.

If Ann and Bob must sit together, we can consider them as one person [AB]. For example, one seating arrangement could be [AB][C][D][E]. Thus, the number of ways to arrange four people in a row is 4! = 24.

However, we must also account for the ways we can arrange Ann and Bob, that is, either [AB] or [BA]. Thus, there are 2! = 2 ways to arrange Ann and Bob.

Therefore, the total number of seating arrangements is 24 x 2 = 48 if Ann and Bob DO sit next to each other.

Since there are 5 people being arranged, the total number of possible arrangements is 5! = 120.

Thus, the number of arrangements when Bob does NOT sit next to Ann is 120 - 48 = 72.

Answer: D
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated &nbs [#permalink] 21 Jul 2018, 09:21
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